Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

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For reaction $ Y_2 + 2Z \rightarrow Product $ , rate controlling step is $ Y + ½ Z \rightarrow Q $ . If the concentration of Z is doubled, the rate of reaction will be

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Explanation

Becomes 1.414 times $ Rate = K [ Y ] [ Z ] ^ { 1 /2 } $ $ \therefore New rate =\sqrt 2 . k[Y] [Z] ^ {1 /2 } =1.414 K [Y] [Z] ^ {1/2 } $

The time for half lif of a certain reaction $ A \rightarrow Products $ , is one hour. When the initial concentration of the reactant A is $ 2 mol L^{-1} $ how much time does it take for its concentration to come from $ 0.50 to 0.25 mole L^{-1} $ if it is a zero order reaction ?

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Explanation

0.25 h For Zero order reaction $ K = { [A] _o \over 2 t { 1/ 2 }} = { 2 \over 2 \times 1 } = 1 mol L^ { -1} hr^{-1} $ $ t = { [ A]_o - [A] \over K } = { 0.50 - 0.25 \over 1 } = 0.25 hr $

For a first order reaction $ A \rightarrow Products $ , the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is

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Explanation

$ 3.47 \times 10 ^ {-4} M min^{-1} , K = { 2.303 \over 40 } log { 0.1 \over 0.025 } = 0.03466 min ^ {-1} $ $ Rate = K [A] ^1 = 0.03466 \times 0.01 =3.466 \times 10 ^ {-4} M min ^ {-1} $

In the reaction $ 2N_2O_5 \rightarrow 4NO_2 + O_2 , initial pressure is 500 atm and rate constant K is 3.38 10^{-5} sec^{-1} . After 10 minutes the final pressure of N_2O_5 $ is

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Explanation

490 atm , $ K = { 2.303 \over t} log { Po \over Pt } \therefore 3.38 \times 10 ^ {-5} = { 2.303 \over 600 } log { 500 \over Pt } $ $ log { 500 \over Pt } = 0.0088 OR { 500 \over pt } = 1.021 OR pt = 490 atm $

The rate constants $K_1 and K_2 for two different reactions are 10^{16}.e^{-2000/T} and 10^{15} .e^{-1000/T} $ respectively. The temperature at which $ K_1 = K_2 $ is

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Explanation

$ { 1000 \over 2.303 } k , k_1 = k_2 $ $ \therefore 10 ^ {16} .e^{-2000 / T } = 10 ^ {15 } .e^{-1000 / T } $ $ \therefore 10.e^{-2000 / T } = 1 .e ^ {-1000 / T } $ $ \therefore ln 10 - {2000 \over T } = - { 1000 \over T } $ $ \therefore 2.303 - { 2000 \over T } = - {1000 \over T } $ $ \therefore T = - { 1000 \over 2.303 } K $

Which catalyst forms $ NH_3 and CO_2 $ from urea ?

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Which of the following is an example of surface catalysis ?

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Which catalyst is used in inversion of sucrose ?

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Which catalyst is used to obtain methanal from water gas ?

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Which catalyst is used in the decomposition of ozone ?

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