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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.
| 1901. |
The conversion of `A rarr B` follows second-order kinetics. Doubling the concentration of `A` will increase the rate of formation of `B` by a factor |
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Answer» Correct Answer - 4 `Ararr B` follows `II` order `:.` Rate `r_(1)=K[A]^(2) …(1)` Now if concentration of A is doubled So rate, `r_(2)=K[2A]^(2) …(2)` Compare the eqs. `(1)` and `(2)` `r_(1)/r_(2)=1/4` or `r_(2) =4r_(1)` `:.` Rate become `4` times |
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| 1902. |
The conversion of `A rarr B` follows second-order kinetics. Doubling the concentration of `A` will increase the rate of formation of `B` by a factorA. `1//4`B. 2C. `1//2`D. 4 |
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Answer» Correct Answer - d `r = k (A)^(2)` , when concentration is doubled `r = k (2A)^(2) = k 4(A)^(2)` the rate becomes 4 times . |
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| 1903. |
The conversion of `A rarr B` follows second-order kinetics. Doubling the concentration of `A` will increase the rate of formation of `B` by a factorA. `1//4`B. `2`C. `1//2`D. `4` |
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Answer» Correct Answer - D `r = k(A)^(2)`, when concentration is doubled `r = k(2A)^(2) = k4(A)^(2)` the rate becomes `4` times. |
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| 1904. |
For an exothermic chemical process occurring in two as (i) `A + B rarr X` (slow) (ii) `X rarr AB` (fast) The process of the reaction can be best described byA. B. C. D. None of these |
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Answer» Correct Answer - 1 Note that after formation of `X (` a slow process ) the decomposition of X is fast. The energy of activation for formation of X is high. |
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| 1905. |
Inversion of sucrose `(C_(12)H_(22)O_(11))` is first-order reaction and is studied by measuring angle of rotation at different instants of time `underset(d)underset("Sucrose")(C_(12)H_(22)O_(11))+H_(2)Ooverset(H^(+))rarrunderset(d)underset("Glucose")(C_(6)H_(12)O_(6))+underset(l)underset("Fructose")(C_(6)H_(12)O_(6))` If `(r_(oo)-r_(0))=a` and `(r_(oo)-r_(t))=(a-x)` (where `r_(0),r_(t)` and `r_(oo)`) are the angle of rotation at the start, at the time t and at the end of the reaction respectively, then there is `50%` inversion when :A. `r_(0)=2r_(t)-r_(oo)`B. `r_(0)-r_(t)-r_(oo)`C. `r_(0)=r_(t)-2r_(oo)`D. `r_(0)=r_(t)+r_(oo_` |
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Answer» Correct Answer - A |
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| 1906. |
Optical rotation of sucrose in 1 N Hcl at various times was found as shown below : `{:("Time (sec)",0,7.18,18.0,27.05,oo),("Rotation (deg)",+24.09,+21.7,+17.7,+15.0,-10.74):}` Shwo that the inversion of scrose is a first order reaction |
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Answer» Initial concentration `[A]_(t)= r_(t)r_(oo)` At time, `[A]_(t)= r_(t)r_(oo)` `k=11.8xx10^(-3)s^(-1)` |
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| 1907. |
For an exothermic chemical process ocuuring in two process occuring in two steps as follows `(i)A+BtoX("slow")" "(ii)XtoAB("fast")` The progress of reaction can be best described by :A. B. C. D. |
| Answer» Correct Answer - A | |
| 1908. |
This exothermic reaction is catalyzed by `MnO_(2)(s)`. `2H_(2)O_(2)(aq) rarr 2H_(2)O(l)+O_(2)(g)` Which of the following will increase the rate of this reaction? (P) Raising the temperature (Q) Increasing the surface area of `MnO_(2)(s)`A. P onlyB. Q onlyC. Both P and QD. Neither P nor Q |
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Answer» Correct Answer - C |
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| 1909. |
The half life period of a first order chemical reaction is `16.93` minutes. Time required for the completion of `99%` of the chemical reaction will be `("log 2"=0.301)`:A. `23.03` minutesB. `112.49` minutesC. `460.6` minutesD. `230.3` minutes |
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Answer» Correct Answer - B |
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| 1910. |
Sucrose isA. Zero order reactionB. First order reactionC. Second order reactionD. Third order reaction |
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Answer» Correct Answer - b Rate = k (Sucrose) `(H_(2)O)^(o)` |
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| 1911. |
For the first order reaction with rate contant `k`, which expression gives the half life period ? (Initail conc. = a)A. `(1^(2))/(k)`B. `(1)/(ka)`C. `(0.693)/(k)`D. `(3)/(2ka^(2))` |
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Answer» Correct Answer - c `t_(1//2) = (0.693)/(k)`. |
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| 1912. |
If the reaction, `A + 2Brarr 3C+D`, which of the following expression does not describe changes in the concentration of various species as a function of time :A. `{(d[C])/(dt)}=-{(3d[A])/(dt)}`B. `{(3d[D])/(dt)}={(d[C])/(dt)}`C. `{(3d[B])/(dr)}=-{(2d[C])/(dt)}`D. `{(2d[B])/(dt)}=-{(d[A])/(dt)}` |
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Answer» Correct Answer - D |
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| 1913. |
The rate law of the reaction `A+2Brarr` Product is given by `(d("Product"))/(dt)= k[A]^(2)[B]`. A is taken in large excess, the order of the reqaction will beA. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - b It is a second order reaction . |
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| 1914. |
For rate of reaction ` A +B+C to ` products is given by ` r = k [A][B]^(0)[C]` , if A is taken in large excess , the order of the reaction would be |
| Answer» Correct Answer - B | |
| 1915. |
The rate law of the reaction `A+2Brarr` Product is given by `(d("Product"))/(dt)= k[A]^(2)[B]`. A is taken in large excess, the order of the reqaction will beA. 3B. 1C. 2D. 0 |
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Answer» Correct Answer - A During the hydrolysis of cane sugar (also called inversion of cane sugar), it concentration of water (the solvent) is so larger that it practically remains constant. Therefore, the inversion of cane sugar is pseudo first order in cane sugar in aqueous solution. Also not that, `[H^(+)]` remains practically constant as it is regenerated in the reaction. |
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| 1916. |
The rate of reaction `A + B rarr` Product is given by the equation `r = k[A][B]`. If `B` is taken in large excess, the order of the reaction would beA. 2B. 1C. 0D. unpredictable |
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Answer» Correct Answer - B b) it is the correct answer as the reaction is pseudo unimolecular reaction. |
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| 1917. |
The rate constant for a reaction is `1.5 xx 10^(-7)` at `50^(@)`C and `4.5 xx 10^(7)s^(-1)` at `100^(@)`C . What is the value of activation energy?A. `2.2 xx 10^(3) J mol^(-1)`B. `2300 J mol^(-1)`C. `2.2 xx 10^(4) J mol^(-1)`D. `220 J mol^(-1)` |
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Answer» Correct Answer - C c) `log k_(2)/k_(1) = E_(a)/(2.303R) (T_(2)-T_(1)]/(T_(1)T_(2))` `(log 4.5 xx 10^(7))/(log 1.5 xx 10^(7)) = E_(a)/(2.303 xx 8.314) xx [(373-323)/(373 xx 323)]` log3 `= (E_(a) xx 50)/(2.303 xx 8.314 xx 19.147 xx 373 xx 323)` `E_(a) = 22011 J mol^(-1)=2.2 xx 10^(4) J mol^(-1)` |
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| 1918. |
According to the collision theory, only the effective collisions among the reacting species result in the products. In order that the collisions may be effective, the reacting species must have energy equal to or more than a certain minimum energy called threshold energy `(E^(@))`. Extra energy which has be supplied to the reactants to make collisions effective is known as activation energy `(E_(a))`. This is related to reaction rate with the help of Arrhenius equation, `k=Ae^(-Ea//RT)` The equation also helps in calculating the activation energy for a reaction at a specific temperature. In general, the rate of reaction is inversely proportional to the activation energy required. The chemical reactions in which the reactants require high amount of activation energy are generally.A. slowB. fastC. instaneousD. spontaneous |
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Answer» Correct Answer - A a) slow reactions require high activation energy. |
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| 1919. |
According to the collision theory, only the effective collisions among the reacting species result in the products. In order that the collisions may be effective, the reacting species must have energy equal to or more than a certain minimum energy called threshold energy `(E^(@))`. Extra energy which has be supplied to the reactants to make collisions effective is known as activation energy `(E_(a))`. This is related to reaction rate with the help of Arrhenius equation, `k=Ae^(-Ea//RT)` The equation also helps in calculating the activation energy for a reaction at a specific temperature. In general, the rate of reaction is inversely proportional to the activation energy required. Which of the following expressions gives the effect of temperature on the reaction rate?A. In k= In A`-E_(a)//RT`B. In k= In A`+E_(a)//RT`C. In k = `A-E_(a)//RT)`D. k= In A`+ "In" E_(a)//RT` |
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Answer» Correct Answer - A a) is the correct answer. |
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| 1920. |
There is no bar on the collisions among the reacting species. Why most of the reactions donot take place under normal conditions? |
| Answer» The reaction species must have energy equal to `E^(@)` and they must be also properly oriented. | |
| 1921. |
According to the collision theory, only the effective collisions among the reacting species result in the products. In order that the collisions may be effective, the reacting species must have energy equal to or more than a certain minimum energy called threshold energy `(E^(@))`. Extra energy which has be supplied to the reactants to make collisions effective is known as activation energy `(E_(a))`. This is related to reaction rate with the help of Arrhenius equation, `k=Ae^(-Ea//RT)` The equation also helps in calculating the activation energy for a reaction at a specific temperature. In general, the rate of reaction is inversely proportional to the activation energy required. The plot of log k Vs `1//T` helps to calculateA. Energy of activationB. Rate constant of the reactionC. Order of reactionD. Energy of activation as well as the frequency factor, |
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Answer» Correct Answer - D d) both energy of activation and frequency factor can be calcualted. |
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| 1922. |
According to the collision theory, only the effective collisions among the reacting species result in the products. In order that the collisions may be effective, the reacting species must have energy equal to or more than a certain minimum energy called threshold energy `(E^(@))`. Extra energy which has be supplied to the reactants to make collisions effective is known as activation energy `(E_(a))`. This is related to reaction rate with the help of Arrhenius equation, `k=Ae^(-Ea//RT)` The equation also helps in calculating the activation energy for a reaction at a specific temperature. In general, the rate of reaction is inversely proportional to the activation energy required. An increase in reaction rate with rise in temperature is due toA. an increase in the number of collisonsB. an increase in the number of activated molecules.C. lowering the activation energyD. shortening of mean free path. |
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Answer» Correct Answer - B b) is the correct answer. |
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| 1923. |
The rates of most reaction double when their temperature is raised from `298K` to `308K`. Calculate their activation energy. |
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Answer» `2.303 "log" K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]` `K_(2)/K_(1)=2, T_(2) =308 K, T_(1)=298K` `:. 2.303 log 2=E_(a)/8.314xx10/(308xx298)` `E_(a)=52.903xx10^(3) J` or `E_(a)=52.903 kJ ` |
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| 1924. |
In the following reaction, how is the rate of appearance of the underlined Product related to the rate of disappearance of the underlined reactant ? `BrO_(3)^(ɵ)(aq) + 5underline(Br)^(ɵ)(aq) + 6H^(o+) (aq) rarr 3underline(Br_(2))(l) + 3H_(2)O(l)`A. `(d[Br])/(d t) = (5)/(3) (-d[Br^(ɵ)])/(d t)`B. `(d[Br_(2)])/(d t) = (-d[Br^(ɵ)])/(d t)`C. `(d[Br_(2)])/(d t) = -(d[Br^(ɵ)])/(d t)`D. `(d[Br_(2)])/(d t) = (3)/(5)(-d[Br^(ɵ)])/(d t)` |
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Answer» Correct Answer - D For reaction: `BrO_(3)^(ɵ)(aq) + 5Br^(ɵ)(aq) + 6H^(o+) (aq) rarr 3Br_(2)(l) + 3H_(2)O(l)` Rate `= (1)/(5) (-d[Br^(ɵ)])/(d t) = (1)/(3) (+d[Br_(2)])/(d t)` `= (d[Br_(2)])/(d t) = (3)/(5) (-d[Br^(ɵ)])/(d t)` |
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| 1925. |
For a reaction `1//2A to 2B`, rate of disappearance of A is related to the rate of appearance of B by the expression:A. `(-d[A])/(dt) = 4(d[B])/(dt)`B. `(-d[A])/(dt) = 1/4(d[B])/(dt)`C. `(-d[A])/(dt) = 1/4 (d[B])/(dt)`D. `-(d[A])/(dt) = (d[B])/(dt)` |
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Answer» Correct Answer - C c) Rate = `1/(1//2)(d[A])/(dt) = 1/2(d[B])/(dt)` `=(-d[A])/(dt) = 1/4(d[B])/(dt)` |
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| 1926. |
For a reaction `1//2A to 2B`, rate of disappearance of A is related to the rate of appearance of B by the expression:A. `-(d[A])/(dt)=(1)/(2)(d[B])/(dt)`B. `-(d[A])/(dt)=(1)/(4)(d[B])/(dt)`C. `-(d[A])/(dt)=(d[B])/(dt)`D. `(d[A])/(dt)=4(d[B])/(dt)` |
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Answer» Correct Answer - B `(1)/(2)Ararr2B` For the given reaction, `-(2d[A])/(dt)=(1)/(2)(d[B])/(dt)="rate of reaction"` Rate of disappearance of A `-(d[A])/(dt)=(1)/(2xx2)(d[B])/(dt)=(1)/(4)(d[B])/(dt)` |
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| 1927. |
The pre-exponential factor for the free radical addition of chloring is `2xx10^(13)s^(-1)`. Find the rate constant of this reaction at `STP`. |
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Answer» Free radical addition reactions have `E_(a)=0` Thus from `K =Ae^(-E_(a)//RT) (if E_(a)=0)` `K=A=2xx10^(13) sec^(-1)` |
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| 1928. |
Which one is not correct for areaction showing a free radical combination?A. Rate constant = Frequency factorB. Energy of activatio is zeroC. Unit of rate constant and frequency factor depends upon the order of reactionD. The rate of reaction depends upon temperature |
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Answer» Correct Answer - d `K=Ae^(-E_(a)//RT)` For free radical combination, `E_(a)=0` `:. K=A` Also `K` and `A` are temperature independent and thus rate `=K[]^(m)`. |
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| 1929. |
For the reaction, `N_(2)+3H_(2) rarr 2NH_(3)` the rate `(d[NH_(3)])/(dt)=2xx10^(-4) M s^(-1)`. Therefore the rate `-(d[N_(2)])/(dt)` is given as:A. `10^(-4) M sec^(-1)`B. `10^(4) M sec^(-1)`C. `10^(-2) M sec^(-1)`D. `10^(-4) sec^(-1) M^(-1)` |
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Answer» Correct Answer - a `(-d[N_(2)])/(dt)=1/2(d[NH_(3)])/(dt)` |
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| 1930. |
In the following reaction, how is the rate of appearance of the underlined Product related to the rate of disappearance of the underlined reactant ? `BrO_(3)^(ɵ)(aq) + 5underline(Br)^(ɵ)(aq) + 6H^(o+) (aq) rarr 3underline(Br_(2))(l) + 3H_(2)O(l)`A. `(d[Br_(2)])/(dt)=(5)/(3)(d[Br^(-)])/(dt)`B. `(d[Br_(2)])/(dt)=(d[Br^(-)])/(dt)`C. `(d[Br_(2)])/(dt)=-(d[Br^(-)])/(dt)`D. `(d[Br_(2)])/(dt)=-(3)/(5)(d[Br^(-)])/(dt)` |
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Answer» Correct Answer - D |
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| 1931. |
`2Aoverset(K)rarr3B+4C` Rate of disappearance of A is `4xx10^(-2)Ms^(-1)`. Find the rate of appearance of B at the same instant.A. `9xx10^(-2)`B. `10^(-2)`C. `4.5xx10^(-2)`D. `6xx10^(-2)` |
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Answer» Correct Answer - D |
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| 1932. |
Assertion: For a free radical combination, `K = A`. Reason: `E_(a)` is zero for free radical combination.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `K = Ae^(-E_(a)//RT)`, if `E_(a) = 0` then `K = A` |
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| 1933. |
Assertion: A plot of `log-log(dX)/(d t)` vs. `log(a-X)` leads to value of `K` as antilog (inercept) and slope equal to order of reaction. Reason: `(dX)/(d t) = K(a-X)^(n)`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `"log"(dX)/(d t)= log K + n log (a-X)` |
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| 1934. |
A first order reaction is carried out with an initial concentration of `10` mol litre and `80%` of the reactant changes into product in `10 sec`. Now if the same reaction is carried out with an intial concentration of `5` mol per litre the percentage of the reactant changing to the product in `10` sec is:A. `40`B. `80`C. `60`D. `50` |
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Answer» Correct Answer - b Time required to complete a definite fraction is independent of initial concentration for I order reaction. |
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| 1935. |
In a first order of reaction the concentration of reactant decreases from `800 "mol"//d m^(3)` to`50 "mol"//d m^(3)` in `2 xx 10^(2) sec`. The rate constant of reaction in `"sec"^(-1)` isA. `2 xx 10^(4)`B. `3.45 xx 10^(-5)`C. `1.386 xx 10^(-2)`D. `2 xx 10^(-4)` |
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Answer» Correct Answer - C `k = (2.303)/(t)"log"_(10)(a)/(a - x), t = 2 xx 102, a = 800, a - x = 50` `k = (2.303)/(2 xx 10^(2)) "log"_(10) (800)/(50) = (2.303)/(2 xx 10^(2)) log_(10) 16` `= (2.303)/(2 xx 10^(2))log_(10) 2^(4) = (2.303)/(2 xx 10^(4)) xx 4 xx 0.301` `= 1.38 xx 10^(-2) s^(-1)` |
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| 1936. |
For the complex, `Ag^(+)+2NH_(3) hArr [Ag(NH_(3))_(2)^(+)]` `((dx)/(dt))=2xx10^(7)L^(2)mol^(-1)[Ag^(+)]NH_(3)]^(2)-1xx10^(-2)s^(-1)` `[Ag(NH_(3)_(2)^(+)]` Hence, ratio of rate constants of the forward and backward reaction is :A. `2xx10^(7)L^(2)mol^(-2)`B. `2xx10^(9)L^(2)mol^(-2)`C. `1xx10^(-2)L^(2)mol^(-2)`D. `0.5xx10^(-9)L^(2)mol^(-2)` |
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Answer» Correct Answer - B |
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| 1937. |
For the reaction, for which the activation energies for forward and backward reactions are same, then:A. `DeltaH=0`B. `DeltaS=0`C. The order is zeroD. None of these |
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Answer» Correct Answer - a `DeltaH=E_(aFR)-E_(aBR)=0` |
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| 1938. |
A consecutive reaction, `A overset(K_(1))(rarr)B overset(K_(2))(rarr)C` is characterised by:A. Maximum in the concentration of `A`B. Maximum in the concentration of `B`C. Maximum in the concentration of `C`D. Highly exothermicity |
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Answer» Correct Answer - b The redioactive equilibrium occurs for concentration of `B`. |
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| 1939. |
Assertion: Every collision of reactant molecule is not successful. Reason: Every collision of reactant molecule with proper orientation is successful one.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C Successful collision to a chemical reaction when the reactant molecules collide with proper orientation and attain threshold energy level. |
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| 1940. |
Rate for a zero order reaction is `2xx10^(-2) mol L^(-1) s^(-1)` . If the concentration of the reactant after 25s is `0.5M`, the initial concentration must have beenA. `12.5M`B. `0.5M`C. `1.0M`D. `1.25M` |
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Answer» Correct Answer - C For a zero order reaction, the rate of the reaction at all concentrations is equal to the rate constant. Integrated rate law for the zeroth order reaction is `K=([R]_(0)-[R]_(t))/(t)` or `[R]_(0)=[R]_(t)+Kt` `=(0.5M)+(2xx10^(-2)Ms^(-1))(25s)` `=(0.5M)+(0.5M)=1.0M` |
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| 1941. |
The rate constant for a zero order reaction is `2xx10^(-2) mol L^(-1) sec^(-1)`. If the concentration of the reactant after `25 sec` is `0.5 M`, the initial concentration must have been:A. `0.5 M`B. `1.25 M`C. `12.5 M`D. `1.0 M` |
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Answer» Correct Answer - d For zero order reaction, `C_(0)-C_(t)=Kt` if `C_(t)=0.5 M, t=25` sec, `K=2xx10^(-2)` then `C_(0)=1.0M` |
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| 1942. |
For firt order reaction, fraction fo colllisions with proper orientation of colloiding molecules is `1xx10^(-2)`, collision frequency is `2xx10^(4)` and fraction of successive collisions is `0.5xx10^(3)`, then the rate constant for the reaction isA. `2.5 xx10^(-4) sec^(-1)`B. `1xx10^(5) sec^(-1)`C. `1xx10^(-9) `mol `dm^(-3)`D. `1xx10^(5)mol dm^(3)sec^(-1)` |
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Answer» Correct Answer - B K=(oriention of colloidal molecules )`xx` (collision frequency )`xx` ( fraction of successive collisions ) `1xx10^(-2) xx2xx10^(4)xx0.5xx10^(3)=1xx10^(5) S^(-1)` |
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| 1943. |
For the reaction between A and B ` A+B to ` product the following rate data were obtained . ` the reaction isA. Zero orderB. first orderC. second orderD. third order |
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Answer» Correct Answer - D Rate `=K[A]^(x)[B]^(y)` for expt , (1) `1.2xx10^(-2)=k[0.1]^(x)[0.2]^(y) ` `(2) 3.6xx10^(-2)=k [ 0.3 ]^(x)[0.2]^(y)` `(3) 4.8xx10^(-2)=k[0.3]^(x)[0.2]^(y)` Dividing II by `I,3 =[3]^(x) therefore x=1` Dividing III by ,I ,4 `=[2]^(y) therefore y=2` order of reaction `x+y=1+2=3` |
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| 1944. |
Initial concentration of reactant of Zero order reaction is 0.04 M, rate constant of the reaction is `1.2xx10^(-3) mol dm^(-3) sec^(-1)` . Hence the half life time isA. `57.75`secB. `16.66 sec `C. `33.33 sec `D. `5.77 sec` |
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Answer» Correct Answer - B `t_(1//2) =([A_(0)])/(2K)=(0.04)/(2xx1.2xx10^(-3))` |
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| 1945. |
when initial concentration of a reactant is dubled in a reaction its half -life period is not affected , the order if the reaction isA. zeroB. firstC. secondD. more than zero but less than first |
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Answer» Correct Answer - B for a zero order reaction `t_(1//2)` is directly proportional to the intial concentration of the reactant `|R|_(0)` `t_(1//2)prop|R|_(0) ` for a first order reaction `k=(2303)/(t) log ""(|R|_(0))/(|R|)` ` at t_(1//2)|R|=(|R|_(0))/(2)` so the above equation becomes `K=(2303)/(t_(1//2))log""(|R|_(0))/([R]_(0)//2)` `t_(1//2)=(2303)/(K)= log ""2= (2303)/(K)xx.3010` `t_(1//2) = (.693)/(K)` i.e., half life period is independent of initial concentration of a reactant . |
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| 1946. |
the half life of the first order reaction is `346.5 ` sec the rate constant for the reaction isA. `2xx10^(-3)sec^(-1)`B. `5xx10^(2) sec^(-1)`C. `2.4xx10^(2) sec^(-1)`D. `1.44xx10^(-3) sec^(-1)` |
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Answer» Correct Answer - A `k=(0.693)/(t_(1//2))=(0.693)/(346.5)=2xx10^(-3)S^(-1)` |
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| 1947. |
The following data were obtained during the first thermal decompoistion of `N_(2)O_(5) (g)` at constant volume. `2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)` `|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}|` Calculate the rate constant.A. `3.39xx10^(-4)S^(-1)`B. `1.39xx10^(-5)S^(-1)`C. `5.45xx10^(-4)S^(-1)`D. `4.91xx10^(-4)S^(-1)` |
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Answer» When t=0 `" "` 0.5 atm At time t `(0.5-2x)""atm" " "2xatm " "xatm"` Total pressure `(p_(t))=P_(N2O5)+ p_(N_2O_5)+P_(O2)` `=0.5-2x+2x+x` `p_(t)=0.5+x` `x=p_(t)-0.5` `p_(N_2O_5)=0.5-2x` `=0.5-2(p_(t)-0.5)=0.5-2p_(t)+1` `p_(N_2O_5)=1.5-2p_(t)` At `t=100_(s),p_(t)=0.512"atm"` `p_(N_2O_5)=1.5-2xx0.512` `=1.5-1.024=0476"atm"` `k=(2.303)/(t)"log"[(p_(0)(N_(2)O_5))/(p_(A)(N_(2)O_(5)))]` `=(2.303)/(100)"log"(0.5)/(0.476)=(2.303)/(100)xx0.02136` `=4.92xx10^(-4)S^(-1)` |
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| 1948. |
The following data were obtained during the first thermal decompoistion of `N_(2)O_(5) (g)` at constant volume. `2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)` `|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}|` Calculate the rate constant. |
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Answer» Let the pressure of `N_(2)O_(5)` (g) decrease by 2x atm . As two moles of `N_(2)O_(5)` decompose to give two moles of `N_(2)O_(4)` (g) and one mole of `O_(2)` (g), the pressure of `N_(2)O_(4)` (g) increases by 2x atm and that of `O_(2)`(g) increases by x atm. `2N_(2)O_(5) (g) to 2N_(2)O_(4)(g) + O_(2) (g)` `{:("Start "t=0," "0.5" atm",0 " atm",0" atm"),("At time t",(0.5-2x)" atm",2x" atm",x" atm"):}` `p_(t)=P_(N_(2)O_(5)+P_(O_(2)))` =`(0.5-2x)+2x+x=0.5+x` `x=P_(t)-0.5` `P_(N_(2)O_(5) =0.5-2x` `=0.5-2(P_(1)-0.5)=1.5-2p_(t)` At `t=100S,P_(1)=0.512` atm `P_(N_(2)O_(5) =1.5-2xx 0.512=0.476 " atm" ` Using equation (4.16) `k=2.303/t"log"P_(i)/P_(A)=2.303/(100s)"log"(0.5atm)/(0.476atm)` = `2.303/(100s)xx0.0216=4.98xx10^(-4)S^(-1)` |
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| 1949. |
A first order reaction is `50%` complete in `30` minutes at `27^(@)C` and in `10` minutes at `47^(@)C`. The rate constant at `47^(@)C` and energy of activation of the reaction in kJ`//`mole will beA. `0.0693, 43.848 kJ mol^(-1)`B. `0.0560,45.621 kJ mol^(-1)`C. `0.0625, 42.926 kJ mol^(-1)`D. `0.0660, 46.189 kJ mol^(-1)` |
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Answer» Correct Answer - A We know that `k = (2.303)/(t) log((a)/(a-x))` Subtituting the values at the two given conditions `k_(27^(@)) = (0.0693)/(30) = 0.0231 , k_(47^(@)) = (0.693)/(10)= 0.0693` We also know that `"log" (k_(2))/(k_(1)) = (2.303)/(t)"log"((a)/(a-x))` or `E_(a) = (2.303 R xx T_(1)T_(2))/(T_(2) - T_(1)) "log"(k_(2))/(k_(1))` `= (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/(320 - 300) xx "log" (0.0693)/(0.231)` `= 43.848 kJ mol^(-1)` |
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| 1950. |
When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction isA. zeroB. firstC. secondD. more than zero but less than first |
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Answer» Correct Answer - B for a zero order reaction `t_(1//2)` is directly proportional to the initial concentration of the reactant ,`[R]_(0)` `t_(1//2)prop[R]_(0)` For a first order reaction , `k=(2.303)/(t)"log"([R]_(0))/([R])` `"at" " t"_(1//2),[R]=[R]_(0)/(2)` So, the above question becomes `k=(2.303)/(t_(1//2))"log"[R]_(0)/([R]_(0)//2)` `t_(1//2=(2.303)/(k)"log"2=(2.303)/(k)xx.3010=(0.693)/(k))` i.e. half-life period is independent of initial concentration of a reactant. |
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