Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

All the above

Activation energy and frequency factor.

$ 3.2 \times 10^{–4} 10k $ rise, the velocity constant becomes nearly double

512 increase of temperature $ n \times 10 $ Increase reaction rate = $ 2 ^ 9 $ $ \triangle T = 100 -10 = 90 = 9 \times 10 \therefore n = 9 $ $ \therefore Increases reaction rate = 2 ^ 9 = 512 $

$ 43.84 Kj mole ^ {-1} log { K_2 \over K_1} = { Ea \over 2.303 R } \left ( { T_2 -T_1 \over T_1 T_2 } \right) $ $ log \left ( { 0.0693 \over 0.0231 } \right) = { Ea \over 2.303 \times 8.3 } \left ( { 320 -300 \over 300 \times 320 } \right) $ $ log 3 = { Ea \over 1.901} \left( { 20 \over 96000} \right) $ Ea =43.84

$ 1.289 times log { K_2 \over K_1 } = { Ea.DT \over 2.303 RT_2 T_1 } = { 9000 \times 5 \over 2.303 \times 2 \times 300 \times 295 } = 0.1104 $ $ log { K_2 \over K_1 } = 0.1104 , { K_2 \over K_1} = 1.289 , K_2 = K_1 \times 1.289 $

$ K_1 = K_2 A.e^{E_1 \over RT } , K_1 = A_1 .e^{-E_1 \over RT } , K_2 = A_2 .e^ {-E_2 \over RT } $
$ {K_1 \over K_2 } = { A_1 \over A_2 } \times e ^ { (E_2 - E_1 ) / RT } = A.e ^ { (2E_1 -E_1)/RT} = A.e^{E_1 / RT } $ $ \therefore K_1 = K_2 .A.e^{E_1 / RT } $

$ { K_1 ^ 1 \over K_1 } \gt { K_2 ^1 \over K_2 } $ $ log { K_1 ^1 \over K_1 } = { E_1 \over 2.303 R } \left[ { T_2 -T_1 \over T_1 T_2 } \right] , log { K_2^1 \over K_2 } = { E_2 \over 2.303 R } [ {T_2 - T_1 \over T_1 T_2 } ] $ $ Since E_1 \gt E_2 $ $ \therefore log { K_1 ^1 \over K_1 } \div log { K_2 ^1 \over K_2 } \gt 1 or { K_1^1 \over K_1 } \gt { K_2 ^1 \over K_2 }$

1 The rate is not depend upon the reactant present in excess

1