Chemistry MCQs for NEET — Practice Questions with Answers

Definition of the rate of reaction

the activation energy of the reaction is large

$ 0.2 \times 10 ^ {-4} $ $ { d [H_2 ] \over dt } = - 0.3 \times 10 ^ {-4} Ms^{-1} $ $ But Rate = - { 1 \over 3 } { d [ H_2 ] \over dt } = + { 1 \over 2 } { d [ NH_3 ] \over dt } $ $ Hence { d [NH_3 ] \over dt } = - { 2 \over 3} { d [H_2 ] \over dt }= - { 2 \over 3 } ( -0.3 \times 10 ^ {-4} ) = 0.2 \times 10 ^ {-4 } $

$ \triangle $ H of reaction

$ 0.012 atm min^{-1} $ $ - { d [N_2 O_4 \over dt } = + { 1 \over 2 } { d [NO_2 ] \over dt }$ $ - { (0.32 - 0.50 ) \over 30 } = 0.006 = { 1 \over 2} {d [NO_2 ] \over dt } $ $ \therefore { d [NO_2] \over dt } = 0.012 atm min^ {-1} $

$ K = K_1 / K_2 $

$ { 1 \over 6 } Rate = k [A] ^ {1/2} [A]^{1/3} [c] ^1 $ $ \therefore Order of reaction = { 1 \over 2 } + { 1 \over 3} + { 1 \over 1 } = { 11 \over 6 } $

$ Rate = K[A][B]^2 $ Keeping [B] constant, [A] is made a 4 times, rate also become 4 times. Hence rate $ \alpha [A] $ Keeping [A] constant, [B] is doubled, rate becomes 4 times. Hence rate $ \alpha [B] ^ 2$ $ \therefore rate = K[A][B]^ 2$

$ Rate = K[A][B] ^ 2 $ Keeping [B] constant, [A] is made a 4 times, rate also become 4 times. Hence $ rate \alpha [A] $
Keeping [A] constant, [B] is doubled, rate becomes 4 times. Hence $ rate \alpha [B]^2 $ $ \therefore rate = K[A][B]^2$

concentration of reactants