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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1651. |
Consider the following statement : 1 The rate of reaction is always proportinal to the concentration of reactants. 2 The order of an elementary chemical reaction step can be determined by examing its stoichimoetry. 3 The first order reactions follows an exponential time course. Of these statements :A. 1,2 and 3 are correctB. 1 and 2 are correctC. 2 and 3 are correctD. 1 and 3 are correct |
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Answer» Correct Answer - C Statement (1) cannot be correct because the rate of zero order reactions does not depend on concentrartion of reactant |
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| 1652. |
when ammonia is treated with `O_(2)` at elevated temperature the rate of disappearance of ammonia is found to be `3.5 xx10^(-2)` mol `L^(-1) sec (-1)` during a measured time interval. The rate of appearance of water will beA. `3.5xx10^(-2)mol L^(-1)sec^(-1)`B. `5.3xx10^(-2) mol L ^(-1) s ^(-1)`C. `2.1 xx10^(1) mol L^(-1) sec ^(-1) `D. `1.4 xx 10^(-1) mol L^(-1) s^(-1)` |
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Answer» Correct Answer - B Since the relative rates of appearance of the products are governed by the coefficients of the chemical equation, the first step is to write the equation for the reaction: `4NH_(3)(g)+5O_(2)(g)rarr4NO(g)+6H_(2)O(g)` Notice that if `4mol` of `NH_(3)` disappears, then `6mol` of `H_(2)O` from or when `1mol` of `NH_(3)` disappease, then `6//4` or `1.5mol` of `H_(2)O` from. Thus, the rate of appearance of `H_(2)O` is `1.5` times as fast as the rate of disappearance of `NH_(3)` : `(Delta[H_(2)O])/(Deltat)=1.5((Delta[NH_(3)])/(Deltat))` `=(1.5)(3.5xx10^(-2molL^(-1)s^(-1))` `=5.25xx10^(-2)molL^(-1)s^(-1)` `=5.3xx10^(-2)molL^(-1)s^(-1)` |
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| 1653. |
Findingf concentration and time : The rate constant for the decomposition of gasous `N_(2)O_(5) at 55^(@)C is 1.7xx10^(-3)s^(-1)` . If the initial concentration of `N_(2)O_(5)` after five half-lives ? How long will it take for the `N_(2)O_(5)` concentration to fall to `12.5%` of its initial values ? Strategy : Because the unit of rate constant is `time^(-1)` , the decomposition of `N_(2)O_(5)` is a first order reaqction. To find `[N_(2)O_(5)]` after n half lives, multiply its initial concentration by `(1//2)^(n)` since `[N_(2)O_(5)]` drops by a factor of 2 during each successive haslf-life. |
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Answer» After five half-lives `(t=5t_(1//2)), [N_(2)O_(5)]` will bw `(1//2)^(5) = (1//32)` of its initial value. Therefore `[N_(2)O_(5)]= (0.020M)/(32) = 0.00062M` Since `12.5%` of the initial concentration corresponds to `1//8 or (1//2)^(3)` of the initial concentration, the time required is three half-lives: `t=3t_(1//2)=3((0.693)/(k))` `=3((0.693)/(1.7xx10^(-3)s^(-1)))` `= 3(4.1xx10^(2)s)` `= 1.2xx10^(3)s(20 min)` |
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| 1654. |
Using the Arrhenius equation : The rate constant fot the formation of hydrogen iodide from the elemewnts `H_(2)(g)+I_(2)(g)rarr2HI(g)` is `2.7xx10^(-4)L//(mol.s)at 600 k and 3.5xx10^(-3)L//(mol.s) at 650 k` . (a) Find the activation energy `E_(a)` . (b) Calculate the rate constant at 700 k . Strategy : (a) Substitute the data given in the problem statement into the Equation `(4.37)` noted just before this example, then solve for `E_(a)` . , (b) Use the same equation, but substitute for `k_(1), T_(1), T_(2)` and `E_(a)` obtained in (a) and solve for `k_(2)` . |
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Answer» According to equation `(4.37)` `log ((k_(2))/(k_(1))) = ((-E_(a))/(2.303 R)) ((1)/(T_(2)) - (1)/(T_(1)))` or `log ((k_(2))/(k_(1))) = ((E_(a))/(2.303 R)) ((1)/(T_(1)) - (1)/(T_(2)))` Substituting the data gives `log ((3.5xx10^(-3))/(2.7xx10^(-4))) = (E_(a))/(2.303xx8.314 J//(mol.k)) ((1)/(600k) - (1)/(650 k))` `log (1.30xx10^(-1)) = 1.11= (E_(a))/(2.303xx8.314 J mol^(-1)) xx (1.28xx10^(-4))` Hence, `E_(a) = ((1.11)(2.303)(8.314 J mol^(-1)))/(1.28xx10^(-4))` `= 16.6xx10^(4)J mol^(-1)` `= 1.66xx10^(5) J mol^(-1)` (b) Substitute `E_(a) = 1.66xx10^(5)J mol^(-1)` `k_(1) = 2.7xx10^(-5) L//(mol.s)` `k_(2) = "unkonown as yet" , T_(1) = 60 k` we get `log (k_(2))/(2.7xx10^(-4)L//(mol.s)) = (1.66xx10^(5)J mol^(-1))/(2.303xx8.314 J mol^(-1) k^(-1)) ((1)/(600 k) - (1)/(700 k))` `=2.07` Taking antilogarithm `((k_(2))/(2.7xx10^(-4)L//(mol.s)))= 10^(2.07) = 1.2xx10^(2)` Hence, `k_(2) = (1.2xx10^(2))(2.7xx10^(-4))L//mol.s` `=3.2xx10^(-2)L//(mol.s)` |
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| 1655. |
Rate constant of a reaction `(k)` is `175 "litre"^(2) "mol"^(-2)sec^(-1)`. What is the order of reaction ?A. FirstB. SecondC. ThirdD. Zero |
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Answer» Unit of `[("litre")/("mol")]^(n-1)xxsec^(-1)" "....(i)` Given unit of k = `"litre"^(2)"mol"^(-1)"sec"^(-1)" "....(ii)` n-1=2 n=3 |
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| 1656. |
At 500 K, the half-life period of a gaseous reaction at the initial pressure of 80 kPa is 350 sec. When the pressure is 40 kPa, the half life period is 175 sec. The order of reaction isA. zeroB. oneC. twoD. three |
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Answer» `((t_(1//2))_(1))/((t_(1//2))_(2))=((p_(2))/(p_1))^(n-1)` `(350)/(175)=((40)/(80))^(n-1)` `2=((1)/(2))^(n-1)` n-1=-1 n=0 (zero order reaction) |
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| 1657. |
When a reaction is progressingA. The rate of the reaction goes on increasingB. The concentration of the products goes on decreasingC. The concentration of the reactants goes on decreasingD. The reaction rate always remain constant |
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Answer» Correct Answer - c As reaction progresses the concentration of the reactants decreases and the concentration of the product increases . |
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| 1658. |
which one of the following statements is not correct ?A. Catalyst does not initiate any reactionB. the value of equilibrium constant is changed in the presence of a catalyst in the reaction equilibriumC. Enzymes catalyse mainly biochemical reactionD. Coenzymes increase the catalytic activity of enzyme |
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Answer» Correct Answer - B for a reversible reaction , it accelerates the speed of forward as well as backward reaction to the equilibrium i.e., does not charge the equilibrium constant of the reaction but helps to attain the equilibrium faster . |
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| 1659. |
Time required for completion of ionic reactions in comparison to molecular reaction isA. MaximumB. MinimumC. EqualD. None |
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Answer» Correct Answer - b Ionic reactions are very fast reactions i.e. takes place instantaneously . |
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| 1660. |
The reactions with low activation energy are alwaysA. AdiabaticB. SlowC. Non-spontaneousD. Fast |
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Answer» Correct Answer - d Less is the activation energy is the reaction or greater is the activation energy slower is the reaction . |
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| 1661. |
The temperature coefficient of a reaction isA. Specific reaction rate at `25^(@)C`B. Rate of the reaction at `100^(@)C`C. Ratio of the rate constants at temperature `35^(@)C` and `25^(@)C`D. Ratio of the rate constants at two temperature differing by `1^(@)C ` |
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Answer» Correct Answer - c Temperature coefficient `(K_(35^(@)C))/(K_(25^(@)C)) = (K_(308K))/(K_(298K)) = 2` and 3 for most reactions . |
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| 1662. |
Mechanism of a hypothetical reaction `X_(2) + Y_(2) rarr 2XY` is given below: (i) `X_(2) rarr X + X` (fast) (ii) `X+Y_(2) hArr XY+Y` (slow) (iii) `X + Y rarr XY` (fast) The overall order of the reaction will be :A. 1B. 2C. 0D. `1.5` |
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Answer» Correct Answer - D we know that , slowest step is the rate determining step . `therefore "rater "( I) = K_(1)[X][Y_(2)]` Now, from equation (i) i.e., `X_(2) to 2x["fast"]` `K_(eq)=([X]^(2))/([X_(2)])` `[X]= {K_(eq)|X_(2)|}^(1//2).. . (ii) ` Now , substitutethe value of [x] from equation (ii) in equation (i) , we get Rate (I) `=K_(1)(K_(eq))^(1//2)[X_(2)]^(1//2)[Y_(2)]=K[X_(2)]^(1//2)[Y_(2)]` `therefore ` order of reaction `=(1)/(2) +1=(3)/(2)=1.5` |
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| 1663. |
Activation energy `=` threshold energy -…………….. . |
| Answer» Correct Answer - average kinetic energy | |
| 1664. |
Assertion : Instantaneous rate of reaction is equal to dx/dt . Reason : It is the rate of reaction at any particular instant of timeA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion and reason both are false . |
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Answer» Correct Answer - b Instantaneous rate of reaction is equal to small change ini concentration (dx) during a small interval of time (dt) at that particular instant of time divided by the time interval . |
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| 1665. |
Mechanism of hypothetical reaction `X_(2) + Y_(2) to 2 XY` is given below (i) `X_(2) to X + X ` (fast ) (ii) `X + Y_(2) hArr XY + Y` (slow) (iii) ` X + Y to XY` (fast) The overall order of the reaction will beA. 2B. 0C. 1.5D. 1 |
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Answer» Correct Answer - c According to law of mass action r = `K [X] [Y_(2)] " " … (i)` From fast step `K_(eq) = ([X]^(2))/([X_(2)])` `[X]^(2) = K_(eq) [X_(2)]` [X] = `sqrt(K_(eq))[X_(2)]^(1//2) …. (ii)` From equation (i) & (ii) `r = K [X_(2)]^(1//2) [Y_(2)]` `r = K[X_(2)]^(1//2) [Y_(2)]` . Overall order of reaction = 1 + 0.5 = 1.5 |
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| 1666. |
fill up the following with suitable terms (i) Activation energy = Threshold energy - _____ (ii) Half-life period of zero order reaction =____ (iii) Average rate of reaction = ___ (iv) Instantaneous rate of reactionA. `{:("(i)",(ii),(iii),(iv)),("potential energy",(0.693)/(k),(dx)/(dt),(Delta[A])/(Deltat)):}`B. `{:("(i)",(ii),(iii),(iv)),("Energy of reactants",(1)/(k),(Delta[A])/(Deltat),(dx)/(dt)):}`C. `{:("(i)",(ii),(iii),(iv)),("Energy of reaction ",(logk)/(t),(Delta[A])/(Deltat),(dx)/(dt)):}`D. `{:("(i)",(ii),(iii),(iv)),("Average kinetic energy of reactants ",(a)/(2k),(Delta[A])/(Deltat),(dx)/(dt)):}` |
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Answer» Correct Answer - D (I ) Activation energy = Threshold energy - Average energy of reactants (ii) `T_(1//2) ` for zero order reaction `=(a) /(2K)` (iii) Average rate of reaction `=(Delta [A])/(delta t) ` (iv ) instantaneous rate of reaction `=(dx )/(dt) ` |
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| 1667. |
The minus sign in rate `= - (d[A])/(dt)` indicates the _______ in concentration of the ______ with time. The rate of a reaction is always _______quantity. The rate of reaction increases with ______in concentration of reactants. The blanks in the question corresponds toA. decrease, products, positive, increaseB. increase, reactants, negative, decraseC. decrease, reactants, positive, increaseD. increase, porducts, positive, increase |
| Answer» Correct Answer - C | |
| 1668. |
Assertion : The rate of reaction is always negative . Reason : Minus sign used in expressing the rate shows that concentration of product is decreasing .A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion and reason both are false . |
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Answer» Correct Answer - d The rate reaction is never negative . Minus sign used in expressing the rate only shows that the concentration of the reactant is decreasing . |
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| 1669. |
The minus sign in rate `= - (d[A])/(dt)` indicates the _______ in concentration of the ______ with time. The rate of a reaction is always _______quantity. The rate of reaction increases with ______in concentration of reactants. The blanks in the question corresponds toA. decrease, products ,postive,increseB. increase, reactants m negative , decreaseC. decrease , reactants , positive , increaseD. increase , products ,positive , increase |
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Answer» Correct Answer - C The concentration of reaction decreases with time . The rate of reaction is a positive quantity ,The rate of reaction increases with increase in concentration of reactants. |
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| 1670. |
Fill in the blanks in the following table for the reaction `X+Y to Z` , the reaction is of first order w.r.t X and zero order w,r,t Y A. `A=0.2 mol L^(-1) ,B=8xx10^(-2) mol L^(-1)S^(-1),C=0.1 mol L^(-1)`B. `A=0.4mol L^(-1),B=4xx10^(-2)mol L^(-1)S^(-1), C=0.2 mol L^(-1)`C. `A=0.2mol L^(-1),B=2xx10^(-2)mol L^(-1)S^(-1),C=0.4 Mol L^(-1)`D. `A=0.4 mol L^(-1),B=2xx10^*-2) mol L^(-1)S^(-1),C=0.4mol L^(-1)` |
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Answer» Correct Answer - A Rate `=k [x] [Y]^(0)` rate is indepent of the conc of Y and it depends only on the conc of X and it is the first order reaction . from exp `.(1) 2xx10^(-2)=K (0.1)` from exp . `(2) ,4xx10^(-2)=k(A)` Dividing (ii ) and (i) `,(4xx10^(-2))/(2xx10^(-2))=(k(A))/(k(0.1))=(A) /(0.1)` `implies 2xx0.1 =A ` ` implies A= 0.2 mol L^(-1) ` from exp .(3) `B= k (0.4)` Dividing (iii ) and (i ) `(B )/(2xx10^(-1))=(k(0.4))/(k (0.1))=4` `implies B=4xx2xx10^(-2) = 8 xx10^(-2) Mol L ^(-1) S^(-1) ` from exp . (4) `2xx10^(-2)=k(C )` Dividing (iv ) and (i ) ,`(2xx10^(-2))/(2xx10^(-2))=(k(C ))/(K (0.1))=(C )/(0.1)` `implies C= 0.1 mol L^(-1) ` |
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| 1671. |
For a reaction `A_(2)+B_(2) hArr 2AB ` the figure shows the path of the reaction in absence and presence of a catalyst what will be the energy of activation for forward `(E_(f))` and backward `(E_(b))` reaction in presence of a catalyst and `Delta H ` for the reaction gt the dotted curve is the path of reaction in presence of a catalyst . A. `E_(F) = 60 KJ // mol , E_(b) = 70 KJ //mol ,Delta H=20KJ //Mol `B. `E_(f) = 20 KJ //mol ,E_(b)=20 KJ //mol ,Delta H=50KJ //mol `C. `E_(f)=70KJ//mol ,E_(b)=20KJ //mol , Delta H=-10KJ//mol `D. `E_(f)=10KJ//mol ,E_(b)=0 KJ//Mol, Delta H=-10KJ //mol` |
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Answer» Correct Answer - D `E_(f) = 70 -60= 10 KJ // mol ` `E_(b) = 70 -50 = 20 KJ //mol ` ` Delta H= 50-60=- 10KJ //mol ` |
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| 1672. |
The rate of a gaseous reaction is given by the expresison `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is suddenly reduced to one-half of the initial volume. The reaction rate relative to the original rate will beA. `1//24`B. `1//32`C. 32D. 24 |
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Answer» Rate `=K[A]^(2)[B]^(3)` When volume is halved, the concentration will become double. `:. "Rate"=K[2A]^(2)[2B]^(3)` `=32 k=K[A]^(2)[B]^(3)=32xx` Original rate from (i) |
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| 1673. |
Rate of a reaction `A + B rarr` Product, is given as a function of different initial concentration of `A` and `B`. `|{:([A] (mol L^(-1)),(B) (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` Determine the order of the reaction with respect to `A` and with respect to `B`. What is the half life of `A` in the reaction ? |
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Answer» Correct Answer - `t_(1//2) = 1.386 min` `OR` w.r.t. `A = 1` `OR` w.r.t. `B = 0` `|{:([A] (mol L^(-1)),[B] (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` form careful observation of the given data, it can be conclude that doubling the concentration of `A` alone doubles the rate of the reaction, therefore order with respect to `A` is one. Doubling the concentration of `B` alone does not affect the rate of the reaction, therefore order with respect to `B` is zero. Rate `= k[A][B]^(0) = k[A]` or `k = (0.005)/(0.01) = 0.5 min^(-1)` Half life `= (0.693)/(k) = (0.693)/(0.5) = 1.386 min`Correct Answer - `t_(1//2) = 1.386 min` `OR` w.r.t. `A = 1` `OR` w.r.t. `B = 0` `|{:([A] (mol L^(-1)),[B] (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` form careful observation of the given data, it can be conclude that doubling the concentration of `A` alone doubles the rate of the reaction, therefore order with respect to `A` is one. Doubling the concentration of `B` alone does not affect the rate of the reaction, therefore order with respect to `B` is zero. Rate `= k[A][B]^(0) = k[A]` or `k = (0.005)/(0.01) = 0.5 min^(-1)` Half life `= (0.693)/(k) = (0.693)/(0.5) = 1.386 min` |
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| 1674. |
For a reaction `A_(2) + B_(2) hArr 2AB` the figure shows the path of the reaction in absence and presence of a catalyst. What will be energy of activation for forward `(E_(f))` and backward `(E_(b))` reaction in presence of a catalyst and `Delta H` for the reaction? The dotted curve is path of reaction in presence of a catalyst. A. `E_(f) = 60 kJ// mol, E_(b) = 70 kJ // mol, Delta H = 20 kJ// mol`B. `E_(f) = 20 kJ mol, E_(b) = 2- kJ// mol, Delta H = 50 kJ//mol`C. `E_(f) = 70 kJ//mol, E_(b) = 20 kJ//mol, Delta H = 10 kJ//mol`D. `E_(f) = 10 kJ//mol, E_(b) = 20 kJ//mol, Delta H = - 10 kJ mol` |
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Answer» Correct Answer - D `E_(f) = 70 - 60 = 10 `kj/mol `E_(b) = 70 - 50 = 20` kj/mol `Delta H = 50 - 60 = - 10` kj/mol |
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| 1675. |
A certain reactions is `50%` complete in 20 minutes at 300 K and the same reaction is again 50% completely in 5 minutes at 350 K. Calculate the activation energy if the reactions is of first order. |
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Answer» `t_(1//2) = 0.693/k` or `k =0.693/t_(1//2)` `k_(1) = 0.693/20 min= 0.03465 min^(-1)` and `k_(2) = 0.693/(5"min") = 0.1386 min^(-1)` According to Arrhenius equation, `logk_(2)/k_(1) = (E_(a))/(2.303R) [1/T_(1)-1/T_(2)]` Hence, `k_(1)= 0.03465 min^(-1)` , `k_(2)=0.1386 min^(-1), T_(1) = 300K, T_(2)=350K` `log 4= (E_(a) xx 50)/(2.303 xx (8.314 Jmol^(-1) K^(-1))xx300 xx 350)` `0.6020 = E_(a)/(2.303 xx (8.314 J mol^(-1)) xx (2100))` `E_(a) = 0.6020 xx 2.303 xx (8.314 J mol^(-1)) xx (2100)` `=24205.8 J mol^(-1) = 24.2 kJmol^(-1)` |
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| 1676. |
For a gaseous reaction: `A(g) rarr B(g)`, the rate expresison may be given asA. `-(d[A])/(dt) = k[A]^(n)`B. `-(1)/(V)(dn_(A))/(dt) = k[A]^(n)`C. `-(1)/(RT)(dP_(A))/(d t) =k[A]^(n)`D. `-(dP)/(dt) = k[P_(A)]^(n)` |
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Answer» Correct Answer - A::B::C::D Rate of reaction `= (-d[A])/(dt) = k[A]^(n)` or `(dP_(A))/(dt) = k(P_(A))^(n)` `= (1)/(V)(-dn_(A))/(dt) = k[A]^(n)` , `[because [A](n_(A))/(V)]` `= -(1)/(RT)(-dP_(A))/(dt)=k[A]^(n)` , `[because [A](P_(A))/(RT)]` |
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| 1677. |
For a first order reaction, the rate of the reaction doubled as the concentration of the reactant is doubled. |
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Answer» Correct Answer - T For a first order reaction, the rate of the reaction doubled as the concentration of the reactant is doubled because the rate is directly proportional to the concentration of the reactants.Correct Answer - T For a first order reaction, the rate of the reaction doubled as the concentration of the reactant is doubled because the rate is directly proportional to the concentration of the reactants. |
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| 1678. |
Select the correct statement `(s)`.A. The order of a reaction may be changed with change in the experimental conditions.B. The rate of reaction, either exotherimc or endothermic, both decreases with decrease in the temperature.C. A reaction mixture thermofyanmically stable should be kinetically unstable.D. A negative catalyst increases the energy of activation. |
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Answer» Correct Answer - A::B Kinetics stability and thermofynamic stability has no relation. Also, the function of negative catalyst to slow down the speed of reaction is not at all related with energy of activation. |
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| 1679. |
In the Arrhenius equation, `k = A exp^(-Ea//RT)`, A may be termed as the rate constant at…………. . |
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Answer» Correct Answer - zero activation energy, infinite temperature In the Arrhenius equation, `k = Ae^(-E_(a)//RT)`, A may be termed as the rate constant at zero activation energy or infinite temperature. This is because only under the conditions `(-E_(a))/(RT) = 0`Correct Answer - zero activation energy, infinite temperature In the Arrhenius equation, `k = Ae^(-E_(a)//RT)`, A may be termed as the rate constant at zero activation energy or infinite temperature. This is because only under the conditions `(-E_(a))/(RT) = 0` |
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| 1680. |
In Arrhenius equation `k = A exp (-(E_(a))/(RT))`. A may be termed as the rate constant atA. Very low temperatureB. Very high temperatureC. Zero activation energyD. The boiling temperature of the reaction mixture |
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Answer» Correct Answer - B::C `k = Ae^-(Ea//RT)` `k = A`: when `T rarr oo` or `E_(a) = 0` or both |
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| 1681. |
In the Arrhenius equation: `k = A exp(-E_(a)//RT)`, the rate constantA. Decreases with increaisng activation energy and increases with temperature.B. Increases with activation energy and temperature.C. Decreases with activation energy and temperature.D. Increases with activation energy and decreaisng temperature. |
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Answer» Correct Answer - D It `T` is high, if `E_(a)` is high, `k` is low. It `T` is low, `k` is low, if `E_(a)` is low, `k` is high. |
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| 1682. |
`2X(g) rarr 3Y(g)+2Z(g)` `{:("Time (in min)",0,100,200,):}` `|{:("Partial pressure of X (mm Hg)",800,400,200,),("Assuming ideal gas condition, calculate",,,,):}|` (a) Order of reaction (b) Rate constant ( c) Time taken for `75%` completion of reaction (d) Total pressure when `p_(x) = 700 mm` |
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Answer» (i) Data shows that half of the reaction is contant, i.e. 100 min , hecne it is a first order reaction , (ii) `k=(0.693)/(t_(1//2))=(0.693)/(100)=693xx10^(-3)"min"^(-1)` (iii) `t_(75%),i.e.,t_(3//4)=2xxt_(1//2)=2xx100=200min` `2X(g)rarr3Y(g)+2Z(g)` `{:(t=0,800,0,0),(dt,800-2X,3X,2X):}` Total prssure =800-2X+3X+2X=800+3X....(i) 800-2X=700 X=50 `:.` Total pressure `=800+3xx50=950` mm Hg |
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| 1683. |
A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear?A. `6 xx 10^(-8) sec`B. `6 xx 10^(-7) sec`C. `6 xx 10^(-9) sec`D. `6 xx 10^(-10) sec` |
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Answer» Correct Answer - C time Total for drop to disappears `(a_(0) - a_(t)) = k_(t) a_(t) = 0` `(3.0 xx 10^(-6))/((0.05 xx 10^(-3)) xx 1.0 xx 10^(-7)) = t_(100%)` `rArr t_(100%) = 6 xx 10^(-9) sec` Alternative method Units rate constant `= mol L^(-1) sec^(-1) rArr` Zero order reaction. `[A_(0)] - [A] = kT` `[A_(0)] = (3 xx 10^(-6) xx 100)/(0.05 xx 10^(-3)) = (3)/(50)` `(3)/(50) - 0 = kT` `rArr t = (3)/(50) xx (10^(-7))/(1)` `= 6 xx 10^(-9) sec` |
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| 1684. |
For reaction a `A tox` P , when [A] = 2.2 mM , the rate was found to be 2.4 m M `s^(-1)` . On reducing concentration of A to half , the rate changes to 0.6 m M `s^(-1)` . The order of reaction with respect to A isA. 1.5B. `2.0`C. `2.5`D. `3.0` |
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Answer» Correct Answer - b `a A to x`P Rate of constant = `[A]^(a)` Order of reaction = a `[A]_(1) = 2.2` mM , `r_(1) = 2.4 m M s^(-1) " " … (i)` `[A]_(2) = 2.2//2 m M , r_(1) = 0.6 m M s^(-1)` or `(2.4)/(4) " " … (ii)` On reducing the concentration of A to half , the rate of reaction is decreased by four times . Rate of reaction = `[A]^(2)` Order of reaction = 2 . |
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| 1685. |
For a reaction `Rto P`, the concentration of a reactant changes from 0.03 M to 0.02M in 25 minutes. Calculate the average rate of the reaction using the units of seconds.A. `6.66xx10^(-5)`B. `6.6xx10^-6`C. `5.67xx10^(-5)`D. `7.26xx10^(-6)` |
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Answer» Correct Answer - B For the reaction, `RrarrP` Average rate of reaction `=("change in concentration of reactant/product")/("time taken")` `=(Delta[R])/(Deltat)=-[(R_(2)]-[R_1)]/(t_(2)-t_(1))` `((0.02-0.03)M)/((25xx60)s)=((-0.01)M)/((1500)s)` `=6.66xx10^(-6)Ms^(-1)` `=6.66xx10^(-6)molL^(-1)s^(-1)` |
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| 1686. |
A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear?A. `6xx10^(-8)` secB. `6xx10^(-7)` secC. `6xx10^(-9)` secD. `6xx10^(-10)` sec |
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Answer» Correct Answer - C |
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| 1687. |
A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear? |
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Answer» `"Concentrationof drop"=("mole")/("volume in mL")xx1000` `=(3.10^(-6))/(0.05)xx1000 =0.06 "mol litre"^(-1)` `"Rate of disappearance"=("conc.change")/("time")` `1xx10^(7)=(0.06)/("time")` `"Time"=6xx10^(-9)sec` |
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| 1688. |
A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear?A. `6xx10^(-8)` sB. `6 xx 10^(-7)` sC. `6 xx 10^(-9)` sD. `6 xx 10^(-10)` s |
| Answer» Correct Answer - C | |
| 1689. |
For reaction `aA rarr xP`, when `[A] = 2.2 mM`, the rate was found to be `2.4mM s^(-1)`. On reducing concentration of `A` to half, the rate changes to `0.6 mM s^(-1)`. The order of reaction with respect to `A` isA. `1.5`B. `2.0`C. `2.5`D. `3.0` |
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Answer» Correct Answer - B `aA rarr xP` Rate of reaction `= [A]a` Order of reaction `= a` `[A]_(1) = 2.2 mM, r_(1) = 2.4 mM s^(-1)" "`…(i) `[A]_(2) = 2.2//2 mM, r_(1) = 0.6 mM s^(-1) = or, (2.4)/(4)`" "… (ii) On reducing the concentration of `A` to half, the rate of reaction is decreased by four times. Rate of reaction `= [A]^(2)` Order of reaction `= 2` |
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| 1690. |
For reaction `aA rarr x//P`, when `[A] = 2.2 mM`, the rate was found to be `2.4mM s^(-1)`. On reducing concentration of `A` to half, the rate changes to `0.6 mM s^(-1)`. The order of reaction with respect to `A` isA. `1.5`B. `2.0`C. `2.5`D. `3.0` |
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Answer» Correct Answer - B `aA rarr xP` Rate of reaction `1= [A]^(a)` order of rection `= a` `[A]_(1) = 2.2mM, r_(1) = 2.4 mM s^(-1)` …(i) `[A]_(2) = 2.2//2 mM, r_(1) = 0.6 mM s^(-1)` or, `(2.4)/(4)` …(ii) On reducing the concetration of `A` to half, the rate of reaction is decreased by four times. Rate of reaction `= [A]^(2)` Order of reaction `= 2`. |
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| 1691. |
In a particular reduction process, the concentration of a solution that is initially 0.24 M is reduced to 0.12 M in 10 hours and 0.06 M in 20 hours. What is the rate constant for the reaction? |
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Answer» Let us calculate the rate constant (k) in both the cases. Ist case: a=0.24 M, (a-x)=0.24-0.12 = 0.12 M, t=10 hr For the first order reaction, `k=2.303/t log a/(a-x) = (2.303)/(10 hr) log (0.24M)/(0.12M) = 2.303/(10 hr) log 2=(2.303 xx 0.3010)/(10 hr)` `(0.693)/(10hr) = 0.693 hr^(-1) = 6.93 xx 10^(-2) hr^(-1)` IInd case: a=0.24 M, a-x =0.24-0.06=0.18M, t=20 hr. For the first order reaction, `k=(2.303)/t log a/(a-x) = 2.303/(20 hr) log (0.24 M)/(0.06 M) = (2.303)/(20 hr) log4 = (2.303)/(20 hr) xx 2 log 2` `=(2.303 xx 2 xx 0.3010)/(20 hr) = 0.0693 hr^(-1) = 6.93 xx 10^(-2) hr^(-1)` |
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| 1692. |
If the fermentaion of sugar in an enzymatic solution that is `0.12` M, the concentration of the sugar is reduced to `0.06` M in 10 h and to `0.03` M in 20 h. What is the order of the reaction?A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - A |
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| 1693. |
In a reaction with initially `0.12 M`, the concentration of reactant is reduced to `0.06 M` in `10` hour and to `0.03 M` in `20` hour. (i) What is order of reaction? (ii) What is rate constant? (b) The rate of a first order reaction is `0.04 mol litre^(-1) s^(-1)` at `10` minute and `0.30 mol litre^(-1)` at `20` minute after initiation. Find the half-life of the reaction. |
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Answer» Assuming I order, i.e., `K=2.303/t"log"_(10) a/((a-x))` For case `I: a=0.12 M, (a-x)=0.06 M, t=10 hr` `K=2.303/10"log"_(10)0.12/0.06=0.069 hr^(-1)` For Case `II: a=0.12 M, (a-x)=0.03 M, t=20 hr` `K=2.303/20"log"_(10)0.12/0.03=0.069 hr^(-1)` `:.` Reaction is of I order and rate constant `K=0.069 hr^(-1)` (b) Rate `=Kxx[A]` `0.04=K[A]_(10)` and `0.03=K[A]_(20)` `([A]_(10))/([A]_(20))=0.04/0.03=4/3` Also `t=2.303/K"log"([A]_(10))/([B]_(20))` when `t=10 min` `10=2.303/K"log" 4/3` `:. K=2.303/10"log"4/3=0.0288 min^(-1)` `t_(1//2)=0.693/K=0.693/0.0288=24.06 min` |
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| 1694. |
In a first order reaction, the concentration of the reactants is reduced to `25%` in one hour. The half-life period of the reactions isA. 2 hrsB. 2hrsC. `1//2` hrsD. `1//4` hrs |
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Answer» Correct Answer - C `k=(2.030)/(t)log.((a)/(a-x))` `k=(2.303)/(1) log. ((100)/(5))=2xx0.693` `t_(1//2)=(0.693)/(k)=(0.693)/(2xx0.693)=(1)/(2)hr` |
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| 1695. |
A drop of solution `("volume" =0.05 mL)` contains `3xx10^(-2)` mole of `H^(+)`. If the rate constant of disappearance of `H^(+)` is `1.0xx10^(2) mol litre^(-1) minute^(-1)`. How long (in minutes) will it take to disappear all the `H^(+)` ions? |
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Answer» Correct Answer - 6 `0.05 mL` has `3xx10^(-2)` mole of `H^(+)` `1000 mL` has `(3xx10^(-2)xx1000)/(0.05)=600` mole `H^(+)` `K=X/t` (zero order, follow unit of K) `t=600/(1xx10^(2))=6 minute` |
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| 1696. |
The reaction `Cis-XhArr` trans -X 1st order in both directions .At `25^(@)C` , the equlibrium constant is 0.10and the constant `k_(f)=3xx10^(-4)sec^(-1)` In an experiment starting with the cis form , How long will it take for half of the equilibrium amount of the trans to be formed?A. 150 sec.B. 200sec.C. 240 sec.D. 210sec. |
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Answer» Correct Answer - D |
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| 1697. |
For reaction `aArarrxP` when [A] =2.2 m M, the rate was found to be 2.4 m `M s^(-1)` . On reducing concentration of A to half, rate changes to `0.06 m M s^(-1)` . The order of reaction with respect to A is :A. `1.5`B. `2.0`C. `2.5`D. `3.0` |
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Answer» Correct Answer - B Rate =`k=[A]^(a)` `2.4 k[2.2]^(a)..........(i)` `0.6 =k[1.1]^(a).........(ii)` Dividing eq. (i) eq. (ii) , a=2 `:.` order =2 |
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| 1698. |
in a reacton, of acidified hydrogen peroxide with potassium iodide, the concentration of iodine formed rises from 0 to `10^(-5) "mil dm"^(-3)` in 10 seconds. What is the rate of reaction ?A. `10^(-6)"mol dm"^(-3)s^(-1)`B. `10^(6)"mol dm"^(-3)s^(-1)`C. `10^(-5)"mol dm"^(-3)s^(-1)`D. `10^(4)"mol dm"^(-3)s^(-1)` |
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Answer» The reaction is `2I^(-)+H_(2)O_(2)+2H^(+)rarrI_(2)+2H_(2)O` `"Rate"=+(d[I_(2)])/(dt)=(10^(-5))/(10)=10^(-6) "mol dm"^(-3)s^(-1)` |
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| 1699. |
At a given temperature, a first-order reaction has a rate constant of `3.3xx10^(-3)s^(-1)`. How much time is required for the reaction to be `75%` complete?A. 100 sB. 210 sC. 420 sD. 630 s |
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Answer» Correct Answer - C |
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| 1700. |
Consider some facts about decomposition of `H_(2)O_(2).` It is catalysed by iodide ion in acidic medium. It is second order reaction with respect to both `H_(2)O_(2).andI^(-1).` Rate equation of this particular reaction will be `Rate=(-d[H_(2)O_(2)]^(2))/(dt)=k[H_(2)O_(2)]^(2)[I^-]` It completes in two steps and both steps are unimolecular elementary reactions. Which of the above written facts are correct, regarding decomposition of `H_(2)O_(2)?` Choose the correct option.A. I and IIB. II and IIIC. III and IVD. I and Iv |
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Answer» Correct Answer - D Decomposition of `H_(2)O_(2)` `2H_(2)O_(2)overset(1^(-))underset("Alkaline medium")rarr2H_(2)O+O_(2)` Rate equation is rate `(-d[H_(2)O_(2)])/(dt)=k[H_(2)O_(2)][1^(-)]` This reaction is first order w.r.t. both `H_(2)O_(2) and 1^(-).` It makes place in two steps and both are bimolecular elementary reactions. |
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