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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.
| 701. |
The rate of reaction between two `A` and `B` decreases by factor `4` if the concentration of reactant `B` is doubled. The order of this reaction with respect to `B` isA. `-1`B. `-2`C. `1`D. `2` |
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Answer» Correct Answer - B Use: `R = k[B]^(n)`, `(1)/(4)R = k[2B]^(n)`, `4 = ((1)/(2))^(n)` , `4 = 2^(-n)`, `n = -2`. |
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| 702. |
The rate of reaction between two `A` and `B` decreases by factor `4` if the concentration of reactant `B` is doubled. The order of this reaction with respect to `B` isA. `-1`B. `-2`C. `2`D. `1` |
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Answer» Correct Answer - B `r = K[A]^(m)[B]^(n)` `(1)/(4) = 2^(n)` `n = - 2`. |
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| 703. |
Which change will decrease the rate of the reaction between `I_(2)(s)` and `H_(2)(g)`?A. Increasing the partial pressure of `H_(2)(g)`B. Adding the `I_(2)(s)` as one piece rather than as several small lonesC. Heating the reaction mixtureD. Adding a catalyst for the reaction |
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Answer» Correct Answer - B |
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| 704. |
The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled . The order of this reaction with respect to reactant B isA. -1B. -2C. 1D. 2 |
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Answer» Correct Answer - b R = `k[B]^(n) , (1)/(4) R = k [2B]^(n) , 4 = ((1)/(2))^(n) , 4 = 2^(-n) , n = -2`. |
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| 705. |
The rate of reaction between two `A` and `B` decreases by factor `4` if the concentration of reactant `B` is doubled. The order of this reaction with respect to `B` isA. `-1`B. `-2`C. 1D. 2 |
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Answer» Correct Answer - B `A + B to "product"` ` Rate Iprop [A]^(x)[B]^(y)` the rate decreases by a factor 4 if the concentration of reaction B is doubled `(I)/(4)prop[A]^(x)[2B]^(y)` from Eqs(i) and (ii) `4=((1)/(2))^(y)` `y=-2` hence order of reaction with respect to B is -2. |
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| 706. |
The first-order disappearance of a substance has a half-life of `34.0` s. How long does it take for the concentration of that substance to fall to `12.5%` of its initial value?A. 11 sB. 68 sC. 102 sD. 272 s |
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Answer» Correct Answer - C |
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| 707. |
The rate constant in numerically the same for the theee reaction of first, second, and third order. Which reaction should be the fastest and it this true for all ranges of concentrations?A. if `[A] = 1` then `r_(1) = r_(2) = r_(3)`B. if `[A] lt 1` then `r_(1) gt r_(2) gt r_(3)`C. if `[A] gt 1` then `r_(3) gt r_(2) gt r_(1)`D. All |
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Answer» Correct Answer - D Rate law for `{:(Ist" order",IInd" order",III" order",,),(Rate = K[A]^(1),R_(2) = K[A]^(2),R_(3) = K[A]^(3),,):}` then we can say `[A] = 1 r_(1) = r_(2) = r_(3)` `[A] lt 1` then `r_(1) gt r_(2) gt r_(3)` `y[A] gt 1` then `r_(3) gt r_(2) gt r_(1)` |
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| 708. |
What is the half life of the irreversible first order reaction, A `rarr` B, if `75%` of A is converted to B in 60 minutes?A. 30 minutesB. 45 minutesC. 60 minutesD. 80 minutes |
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Answer» Correct Answer - A |
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| 709. |
The rate equation for the reaction `2A+B rarr C` is found to be: `rate = k[A][B]`. The correct statement in relation of this reaction is thatA. The value of `k` is indepedennt of the initial concentration of `A` and `B`.B. `t_(1//2)` is a constant.C. The rate of formation of `C` is twice the rate of disappearnce of `A`.D. The unit of `k` must be `s^(-1)` |
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Answer» Correct Answer - A For second order reaction, (a) Correct statement. (b) `t_(1//2) prop (a)^(-1)` ( c) `(d[C])/(dt) = (-d[A])/(2 dt)` (d) For second order, units of `k` is `L mol^(-1) s^(-1)`. |
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| 710. |
Higher order `(gt3)` reaction are rare due to :A. loss of active species on collisionB. low probaility of simultaneous collision energy of all the reacting speciesC. increase inentropy and activation energy as more molecules are involvedD. shifting of equilibrium towards reactants due to elastic collisions |
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Answer» Correct Answer - B Reaction of higher `(gt3)` are rare due to low probability of simultaneous collision of all the reacting species. |
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| 711. |
For the non-stoichiometric reaction `2A+BrarrC+D` The following kinetic data were obtained in theee separate experiment, all at `98 K` `|{:("Initial concentration (A)","Initial concentration (B)","Initial rate of formation of C" (molL^(-1) s^(-1))),(0.1 M,0.1 M,1.2 xx 10^(-3)),(0.1 M,0.2 M,1.2 xx 10^(-3)),(0.2 M,0.1 M,2.4 xx 10^(-3)):}|` The rate law for the formation of `C` is:A. `(dc)/(dt)=k[A][B]^(2)`B. `(dc)/(dt)=k[A]`C. `(dc)/(dt)=k[A][B]`D. `(dc)/(dt)=k[A]^(2)[B]` |
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Answer» Correct Answer - B Let order with respect to A and B are `alpha and beta` resepecitvely `:. (dc)/(dt)=k[A]^(alpha)[B]^(beta)` `1.2xx10^(-3)=k[0.1]^(alpha)[0.1]^(beta)" ".....(i)` `1.2xx10^(-3)=k[0.1]^(alpha)[0.2]^(beta)" ".....(ii)` `2.4xx10^(-3)=k[0.2]^(alpha)[0.1]^(beta)" ".....(iii)` `{:("Dividing (ii) by (i)":,2^(beta),:.beta=0),("Dividing (ii)by(i)":,2^(alpha)=2,:.alpha1),("Rate law will be ":,,):}` `(dc)/(dt)=k[A]` |
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| 712. |
A first order reaction in `75%` complete is 60 minutes. Find the half life of this reaction. |
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Answer» Step I: Calculation of rate constant. `a=100% (a-x) = 25%, t=60 min`. for first order reaction, `k=(2.303)/k log a/(a-x) = (2.303)/(60"min") log 100/25` `=(2.303)/(60 "min") xx log4 = (2.303 xx 0.6020)/(60 min) = 0.0231 min^(-1)` Step II: Calculation or half life period `(t_(1//2))` `t_(1//2) = 0.693/k = (0.693)/(0.0231 min^(-1))=30 min` |
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| 713. |
For the reaction `A+BrarrC`, the rate law is : Rate=k`[A]^(2)`. Which change(s) will increase the rate of the reaction? (P) Increasing the concentration of A (Q) Increasing the concentration of BA. P onlyB. Q onlyC. Both P and QD. Neither P nor Q |
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Answer» Correct Answer - A |
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| 714. |
For a reaction ,`(d[X])/(d t)` is ewqual toA. `k_(1)(a-x)-k_(2)(a-x)`B. `k_(2)(a-x)-k_(1)(a-x)`C. `k_(1) (a-x)+ k_(2)(a-x)`D. `-k_(1)(a-x)-k_(2)(a-x)` |
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Answer» Correct Answer - C `(-d[a-x])/(d t) = K_(1)[a-x]+K_(2)[a-x]` |
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| 715. |
Choose the correct set of identification. A. `{:(underset(E+S rarrES)(DeltaE" for"),underset(ES rarr EP)(E_(a)" for"),underset("for" S rarr P)(DeltaE_("overall")),underset(EP rarr E+P)(E_(a)" for")):}`B. `{:(underset(E+S rarr ES)(E_(a)" for"),underset(E+S rarr ES)(DeltaE" for"),underset(ES rarr EP)(E_(a)" for"),underset("for "S rarr P)(DeltaE" for")):}`C. `{:(underset(ES rarr EP)(E_(a)" for"),underset(EP rarr E+P)(E_(a)" for"),underset("for "S rarr P)(DeltaE_("overall")),underset(EP rarr E+P)(DeltaE" for")):}`D. `{:(underset(E+S rarr ES)(E_(a)" for"),underset(ES rarr EP)(E_(a)" for"),underset(EP rarr E+P)(E_(a)" for"),underset("for "S rarr P)(DeltaE_("overall"))):}` |
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Answer» Correct Answer - B Factual |
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| 716. |
Half lives of decomposition of `NH_(3)` on the surface of a catalyst for different initial pressure are given as `:` `P("torr")" "200" "300" "500` `t_(1//2)" "10" "15" "25` The order of the reaction is `-`A. 2B. 0C. 1D. `0.5` |
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Answer» Correct Answer - 2 `t_(1//2)prop(1)/(a^(n-1))` `(15)/(10)=((200)/(300))^(n-1)` `(3)/(2)=((2)/(3))^(n-1)` `n-1=-1` `n=0` |
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| 717. |
Decomposition of `H_(2)O_(2)` follows a first order reaction . In fifty minutes the concentration of `H_(2)O_(2)` decreases from `0.5` to `0.125 M` in one such decomposition . When the concentration of `H_(2)O_(2)` reaches from `0.05` M , the rate of formation of `O_(2)` will beA. `6.93 xx 10^(-4) mol "min"^(-1)`B. `2.66 L "min"^(-1)` at STPC. `1.34 xx 10^(-2) mol "min"^(-1)`D. `6.93 xx 10^(-2) mol "min"^(-1)` |
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Answer» Correct Answer - a In 50 minutes , concentration of `H_(2)O_(2)` becomes `(1)/(4)` of initial `implies 2 xx t_(1//2) = 50 ` minutes `" " t_(1//2) = 25` minutes `implies K = (0.693)/(25) ` per minute `r_(H_(2) O_(2)) = (0.693)/(25) xx 0.05 = 1.386 xx 10^(-3)` `2H_(2)O_(2) to 2H_(2)O + O_(2) " " r_(O_(2)) = (1)/(2) xx r_(H_(2)O_(2))` `r_(O_(2)) = 0.693 xx 10^(-3) " " r_(O_(2)) = 6.93 xx 10^(-4)` mol/minute `xx` litre |
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| 718. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0 xx 10^(6) s^(-1)` and `9.2 kJ mol^(-1)`B. `6.0 s^(-1)` and `16.6 kJ mol^(-1)`C. `1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`D. `1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - D `k = Ae^(E_(a)//RT)` or `log k = logA - (E_(a))/(2.303 RT)` `log k = -2000((1)/(T)) + 6.0` On comparison, we get `log A = 6` or `A = 10^(6)s^(-1)` `:. (E_(a))/(2.303R) = 2000` or `E_(a) = 2.303 xx 8.314 xx 2000` `~~ 38.3 kJ mol^(-1)`Correct Answer - D `k = Ae^(E_(a)//RT)` or `log k = logA - (E_(a))/(2.303 RT)` `log k = -2000((1)/(T)) + 6.0` On comparison, we get `log A = 6` or `A = 10^(6)s^(-1)` `:. (E_(a))/(2.303R) = 2000` or `E_(a) = 2.303 xx 8.314 xx 2000` `~~ 38.3 kJ mol^(-1)` |
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| 719. |
Why is the probablity of reaction with molecularity higher than three very rare? |
| Answer» Molecularity of a reaction is defined as the number of the reacting molecules/species that are colliding simulatenously. As the number increases, the chances of their simultaneous collisions decreases. This means that it is very little probable than more than three reacting molecules may collide simultaneously. In order words, the probability of reactions with molecularity more than three is very rare. | |
| 720. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `48.6kJ`B. `58.5 kJ mol^(-1)`C. `60.5 kJ`D. `53.6 kJ mol^(-1)` |
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Answer» Correct Answer - D According to Arrhenius equation log `(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` Substituting the data, we have log `2=(E_(a))/(2.303(8.314JK^(-1)mol^(-1)))((310K-300K)/(300Kxx310K))` `E_(a)=53.6KJ//mole` |
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| 721. |
For any reaction , `A+BtoC+D` Rate law obtained is Rate=k[A][B] Select the correct option(s).A. Raction must be elemenaryB. Reaction must be complexC. Reaction may be elementary or complexD. Reaction can not be complex |
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Answer» Correct Answer - C |
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| 722. |
Higher order `(gt3)` reaction are rare due to :A. increases in entropy and activation energy as more molecules are involved.B. shifting of equilibrium towards reactants energy due to elastic collisionsC. loss of active species on collisionD. low probability of simultaneous collision of all the reacting species |
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Answer» Correct Answer - D The probability that more than three molecules can collide and react simultaneously is very small. Hence, the molecularity greater than three is not observed and higher order reactions are less probable. |
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| 723. |
At a given temperature ,`k_(1)=k_(2)` for the reaction, `A+BhArrC+D` `If [(dx)/(dt)]=k_(1)[A][B]-k_(2)[C][D]` in which set o fthe concentration , the reaction ceases?A. `{:("([A])",([B]),([C]),([D])),("(0.1M)",(0.2M),(0.3M),(0.4M)):}`B. `0.4M " "0.25M" "0.2M" "0.5M`C. `0.2M " "0.2M" "0.3M" "0.2M`D. `0.2M " "0.2M" "0.4M" "0.2M` |
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Answer» Correct Answer - B |
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| 724. |
For a reaction of reversible nature , net rate is `((dx)/(dt))=k_(1)[A]^(2)[B]^(1)-k_(2)[C]`, hence , given reactionis:A. `2A+(1)/(B)hArrC`B. `2A+BhArrC`C. `2AhArrC+B^(-1)`D. none of these |
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Answer» Correct Answer - B |
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| 725. |
Which is correct about zero order reactionA. Rate of reaction depends on decay constantB. Rate of reaction is independent of concentrationC. Unit of rate constant is `"concentration"^(-1)`D. Unit of rate constant is `"concentration"^(-1) "time"^(-1)` |
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Answer» Correct Answer - b For zero order reaction , rate of reaction is independent of concentration R = k `["Reactant"]^(0)` |
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| 726. |
Decay of `"_(92)U^(235)` is ….., order reactionA. zeroB. FirstC. SecondD. Third |
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Answer» Correct Answer - b The radioactive disintegration reactions are of first order because in this rate of disintegration depends on the concentration term of radioactive material only . |
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| 727. |
The half-life of 2 sample are `0.1` and `0.4` seconds . Their respective concentration are 200 and 50 respectively . What is the order of reaction |
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Answer» Correct Answer - b `t_(1//2) prop a^(1-n) implies (0.1)/(0.4) = ((200)^(1-n))/((50)^(1-n)) implies (1)/(4) [(4)/(1)]^(1-n) = [(1)/(4)]^(n-1)` `implies (1)/(4^(1)) = (1)/(4^(n-1)) therefore n - 1 = 1 , n = 2`. |
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| 728. |
In the reaction `2N_(2)O_(5) to 4 NO_(2) + O_(2)` , initial pressure is 500 atm and rate constant K is `3.38 xx 10^(-5) "sec"^(-1)` . After 10 minutes the final pressure of `N_(2)O_(5)` isA. 490 atmB. 250 atmC. 480 atmD. 420 atm |
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Answer» Correct Answer - a `p_(0) = 500 ` atm , `K = (2.303)/(t) "log"_(10) (p_(0))/(p_(t))` `3.38 xx 10^(-5) = (2.303)/(10 xx 60) "log" (500)/(p_(t))` or `0.00880 `= log `(500)/(p_(t)) implies (500)/(1.02) = 490` atm . |
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| 729. |
In a first order of reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the raction has occurred ?A. `1//6` of initial concentrationB. `1//64` of initial concentrationC. `1//12` of initial concentrationD. `1//32` of initial concentration |
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Answer» Correct Answer - B `t_(1//2) = 10` No. of half-life `= (60)/(10) = 6` half-life `C_(t) = (C_(0))/(2)^(n) = (C_(0))/(2)^(6) = ((C_(0))/(64))` |
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| 730. |
The decompostion of `N_(2)O_(5 (g)) to NO_(3(g))` Proceeds as a first order reaction with a half-life period of 30 seconds at a certain temperature . If the initial concentration `[N_(2)O_(5)] = 0.4` M , what is the rate constant of the reactionA. `0.00924 sec^(-1)`B. `0.0231 sec^(-1)`C. `75 sec^(-1)`D. `12 sec^(-1)` |
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Answer» Correct Answer - b For a first order reaction , k `= (0.693)/(t_(1//2)) = (0.693)/(30) = 0.0231 sec^(-1)` |
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| 731. |
In the reaction `Cl_(2) + CH_(4) overset(hv)(to) CH_(3) Cl + HCl` , presence of a small amount of oxygenA. Increases the rate of reaction for a brief period of timeB. Decreases the rate of reaction for a brief period of timeC. Does not affect the rate of reactionD. Completely stops the reaction |
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Answer» Correct Answer - d Presence of small amount of oxygen completely stops the reaction . |
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| 732. |
A foreign substance that increases the speed of a chemical reacting is calledA. InhibitorB. PromoterC. ModeratorD. Catalyst |
| Answer» Correct Answer - D | |
| 733. |
A substance having a half-life period of `30` minutes decomposes according to the first order rate law. The fraction decomposed, the balance remaining after `1.5` hours and time for `60%` decomposition on its doubling the initial concentration will be.A. `87.4, 0.126, 39.7` minB. `80.6, 0.135, 40.8` minC. `90.5, 0.144, 2829` minD. `80.2, 0.135, 26.6 min |
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Answer» Correct Answer - A We know that `K = (0.693)/(t_(1//2)) = (0.693)/(30) = 0.023//"min"^(-1)` Let the initial concentration of the substance `= 100` and the substance decomposed in `1.5` hours `(90 min) = x` then, `a-x = 100-x` Substituting these values in `K = (2.303)/(t) "log"((a)/(a-x))` `rArr 0.0231 = (2.303)/(90) "log" (100)/(100-x)` `"log"(100)/(100-x) = (0.0231 xx 90)/(2.303)` `rArr x = 87.4` So the fraction decomposed in `1.5` hours `= (87.4)/(100) = 0.874` Fraction remaining behind after `1.5` hours `= 1 - 0.874 = 0.126` The time required for `60%` of decomposition `K = (2.303)/(t) "log"((a)/(a-x))` `rArr 0.023 = (2.303)/(t) "log"((100)/(40))` `t = (2.303)/(0.0231) "log"((10)/(4)) = 39.7` minutes Since the reaction is of first order the time required to complex specific fraction is independent of initial concentration (or pressure). Hence `60%` of the reaction will decompose in `39.7` mins in this case also. |
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| 734. |
For a given reaction `3 A + B to C + D` the rate of reaction can be represented byA. `-(1)/(3) (d[A])/(dt) = (-d[B])/(dt) = (+d[C])/(dt) = (+d[D])/(dt)`B. `-(1)/(3) (d[A])/(dt) = (d[C])/(dt) = K[A]^(m) [ B]^(n)`C. `+(1)/(3) (d[A])/(dt) = (-d[C])/(dt) = K[A]^(n) [ B]^(m)`D. None of these |
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Answer» Correct Answer - a `-(1)/(3) (d[A])/(dt) = - (d[B])/(dt) = (+d[C])/(dt) = (+d(D))/(dt)` . |
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| 735. |
The rate of chemical reaction at constant temperature is proportional toA. The amount of products formedB. The product of masses of the reactantsC. The product of the molar concentration of the reactantsD. The mean free path of the reaction |
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Answer» Correct Answer - c The rate of chemical reaction `prop` The product of the molar conc., of the reactants (at constant T). |
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| 736. |
The difference rate law for the reaction `H_(2) +I_(2) to 2HI` isA. `- (d[H_(2)])/(dt) = -(d[I_(2)])/(dt) = + (1)/(2) (d[H])/(dt)`B. ` (d[H_(2)])/(dt) = (d[HI])/(dt) = (1)/(2) (d [HI])/(dt)`C. `(1)/(2) (d[H_(2)])/(dt) = (1)/(2) (d[I_(2)])/(dt) = - (d[HI])/(dt)`D. `-2 (d[H_(2)])/(dt) = - 2 (d[I_(2)])/(dt) = + (d [HI])/(dt)` |
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Answer» Correct Answer - ad Concentration of reactants decreases while concentration of product increases. |
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| 737. |
For the reaction `2 N_(2) O_(5(g)) to 4 NO_(2 (g)) to O_(2(g))` , if concentration of `NO_(2)` in 100 seconds is increased by `5.2 xx 10^(-3)` m . Then the rate of reaction will beA. `1.3 xx 10^(-5) ms^(-1)`B. `5 xx 10^(-4) ms^(-1)`C. `7.6 xx 10^(-4) ms^(-1)`D. `2 xx 10^(-3) ms^(-1)` |
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Answer» Correct Answer - a `2N_(2)O_(5(g)) to 4NO_(2 (g)) + O_(2(g))` Rate of reaction with respect to `NO_(2)` `=(1)/(4) (d[NO_(2)])/(dt) = (1)/(4) xx (5.2 xx 10^(-3))/(100) = 1.3 xx 10^(-5) ms^(-1)`. |
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| 738. |
In the synthesis of ammonia by Haber process , if 60 moles of ammonia is obtained in one hour , then the rate of disapperance of Nitrogen isA. 30 mol/minB. 6 mol/minC. `0.5` mol/minD. 60 mol/min |
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Answer» Correct Answer - c `N_(2) + 3H_(2) = 2NH_(3)` rate = `- (d[N_(2)])/(dt) = -(1)/(3) (d[H_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt)` Rate of disappearance of `N_(2)` `= (1)/(2)` of rate of formation of `NH_(3)` `= (1)/(2) xx (60 "mole")/(1"hour")= (1)/(2) xx (60)/(60)` mole/minute = 0.5 mole/minute . |
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| 739. |
A chemical reaction involves two reacting species. The rate of reaction is directly proportional to the conc. Of one of them and inversely proportional to the concentration of the other. The order of reaction isA. ZeroB. 1C. 2D. 4 |
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Answer» Correct Answer - a The order of reaction is zero . Suppose the following reaction takes place. `A + B to` product `therefore " "` rate = k `[A][B]^(-1)` `therefore " "` order = `1 + (-1) = 0`. |
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| 740. |
For a reaction between A and B the order with respect to A is 2 and the order with respect to B is 3 . The concentration of both A and B are doubled the rate will increase by a factor ofA. 12B. 16C. 32D. 10 |
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Answer» Correct Answer - c Given `"Rate"_(1st) = K[A]^(2) [B]^(3) " " … (i)` `"Rate"_(2nd) = K[2A]^(2) [2B]^(3) " " …. (ii)` On Dividing equation (i) and (ii) we get `("Rate"_(1st) )/("Rate"_(2nd)) = (K[A]^(2)[B]^(3))/(K4[A]^(2)8[B]^(3)) = (1)/(32)` `therefore "Rate"_(2nd) = 32 xx "Rate"_(1st)` |
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| 741. |
Why does the rate of any reaction generally decrease during the course of the reaction? |
| Answer» The rate of any reaction is proportional to the concentration of the reactants. With the progress of the reaction, the concentration decreases gradually and so it s the rate of the reaction. | |
| 742. |
The rate law for a reaction between the substances A and B is given by rate `= K[A]^(n) [B]^(m)`. On doubling the concentration of A and having the concentration of B, the ratio of the new rate to the earlier rate of the reactio will be:A. `(1)/(2^(m)+n)`B. (m+n)C. (n-m)D. `2^((n-m)` |
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Answer» Correct Answer - D `"Initial rate"=k[A]^(n)[B]^(n)` when[A]=[2A] and `[B][(1)/(2)B]`, then `"New rate"=k[2A]^(n)[1//28]^m` On dividing Eq.(ii)by Eq.(i), we get `"New rate"/"Initial rate"=(k[2A]^(n)[1//2B]^(m))/(k[A]^(n)[B}^(m))` `"New rate"/"Initial rate"=(2)^(n)(1//2)^(m)=(2)^(n)(2)^(-m)=2(n-m)` Hence, the ratio of new rate fo the reaction is `2^(n-m).` |
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| 743. |
Which of the following is/are correct about the first order reaction ?A. B. C. Half life depends on temperature.D. Rate constant is directly proportional to temperature. |
| Answer» Correct Answer - A::B::C | |
| 744. |
A reaction is `10%` complete in `5min` and `50%` complete in `25 min`. Which of the following is/are correct?A. a. Order of reaction is one.B. b. Order of reaction is zero.C. c. Reaction will be complete in finite time.D. d. Reaction will be complete in infinite time. |
| Answer» Correct Answer - B::C | |
| 745. |
The correct statement (s) areA. Order a reaction is an experimental property.B. Order may change with change in experimental conditions.C. Molecularity concerns with mechanism while order concerns with kinetics.D. A reaction taking place by bimolecular colliison must always be of second order. |
| Answer» Correct Answer - A::B::C | |
| 746. |
The energy profile diagram for the reaction: `CO(g)+NO_(2)(g) hArr CO_(2)(g)+NO(g)` is given below: The activation energy of the forward reaction isA. `x`B. `y`C. `x+y`D. `x-y` |
| Answer» Correct Answer - A | |
| 747. |
The following data were observed for the following reaction at `25^(@)C, CH_(3)OH + (C_(6)H_(5))_(3)C Clrarr(C_(6)H_(5))_(3)C.OCH_(3)+HCl` Rates `(d[C])/(dt)` in sets I, II, and III are, respectively, (in `M min^(-1)`):A. `{:(I,II,III,),(1.30 xx 10^(-4),2.6 xx 10^(-4),1.02 xx 10^(-3),):}`B. `{:(I,II,III,),(0.033,0.0039,0.0077,):}`C. `{:(I,II,III,),(0.02xx10^(-4),0.04 xx 10^(-4),0.017,):}`D. None of these |
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Answer» Correct Answer - A Concentration of the Product has been given: `((d[C])/(dt))_(I) = (0.0033)/(25) = 1.32 xx 10^(4) M min^(-1)` `((d[C])/(dt))_(II) = 2.6 xx 10^(-4) M min^(-1)` `((d[C])/(dt))_(III) = 1.02 xx 10^(-3) M min^(-1)` |
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| 748. |
The following data were observed for the following reaction at `25^(@)C, CH_(3)OH + (C_(6)H_(5))_(3)C Clrarr(C_(6)H_(5))_(3)C.OCH_(3)+HCl` Rate law of the above experiment isA. `k[A]^(2)[B]`B. `k[A][B]`C. `k[A][B]^(2)`D. `k[A]^(2)[B]^(2)` |
| Answer» Correct Answer - A | |
| 749. |
A first order reaction is found to have a rate constant, `k = 4.2 xx 10^(-12) s^(-1)`. Find the half-life of the reaction.A. `1.26 xx 10^(13)s`B. `1.65 xx 10^(11)s`C. `1.65 xx 10^(-11) s`D. `1.26 xx 10^(-13) s` |
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Answer» Correct Answer - B For the first order reaction `t_(1//2)=(0.693)/(k)=(0.693)/(4.2xx10^(-12)s^(-1))` `=1.65 xx10^(11)s` |
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| 750. |
For a certain reaction involving a single reaction, it is found that `C_(0)sqrt(t_(1//2))` is constant where `C_(0)` is the initial concentration of reactant and `t_(1//2)` is the half-life. The order of reaction is:A. `1`B. `0`C. `2`D. `3` |
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Answer» Correct Answer - d For III order,` t_(t//2) prop 3/(2KC_(0)^(2)` |
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