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$ 2A +2B ---> D + E $ For the reaction following mechanism has been proposed. $A + 2B --> 2C +D (slow) $ $ A + 2C --> E (Fast) $ The rate law expression for the reaction is
$ rate = K [A][B]^2$ Rate of reaction for slowest step
$ A2 + B2 ---> 2 AB $ reaction follow the mechanism as given below (i) $ A_2 --> 2A (fast) $ (ii) $ A + B_2 --> AB + B (slow) $ (iii) $ A + B ---> AB (fast) $ the order of overall reaction is
$ 1.5 From slowest step rate = k [B_2] [A] $ $ From 1 ^ {st} eq . Keq = [A] { 2 / [ A_2] } $ $ \therefore [A] = keq ^ {1 /2} . [A_2 ] ^ { 1 /2 } $ $ rate = K [B_2] keq ^ {1/2} . [A_2] ^ {1/2} = k .keq ^ {1/2} [A_2] ^ { 1 /2 } [ B_2 ] = K^1 [A_2 ]^ {1 /2 } [B_2 ] $
For the reaction $ 2A + B ---> Products $ , reaction rate = $ K [A][B]^ 2$ . Concentration of A is doubled and that of B is halved the rate of reaction will be ...
$ halved\; rate ' = k [A] [B] ^ 2 rate ' = k [2A] [{B \over 1}] ^ 2 $ $ = { 1 \over 2 } k [A] [B] ^ 2 $ $ \therefore x" = { 1 \over 2 } x ' $
In one reaction concentration of reaction A is increased by 16 times, the rate increases only two times. The order of the reaction would be ...
$ { 1 /4 } ( 1) r = k [A] ^ n (2) 2 r = k [16 A] ^ n $ $ 2r = K [A]^n 16 ^ n $ $ { 2r \over r } = { K [A] ^ n 6 ^ n \over K [A] ^ n } \therefore 2 = 16 ^ n \therefore n = { 1 \over 4} $
In the reaction $ A ---> B $ . When the concentration of A is changed from 0.1 M to 1 M, the rate of reaction increases by a factor of 100. The order of reaction with respect to A is ….
2 concentration increased = 10 $ times rate increased = 10^ 2 times $ $ \therefore \; Order = 2 $
For the reaction of $ A + B --> C + D$ , doubling the concentration of both the reactants increases the reaction rate by 8 times and doubling the initial concentration of only B simply doubles the reaction rate. The rate law for the reaction is
$ r = K [A]^2 [B] (i) r = k [A] ^x [B]^y (ii) 8r = k [2A] ^x [2B] ^ y (iii) 2r = k [A] ^x [2B] ^ y $ $ z(iii) \div (i) \cong 2^y = 2 \therefore y = 1 $ $ (ii) \div (i) \cong 2 ^x = 4 \therefore x = 2 $
The unit of rate constant for a zero order reaction is
$ mole litre^ {-1} sec ^ {-1} rate = K [R] ^ n , K = { rate \over [R]^ n } = { M/s \over M^n } n = 0 $ $ K = M^{1-n} S^{-1} \therefore K = M/S $
The rate constant of a reaction has same units as the rate of reaction. The reaction is of ...
$ zero order K = { rate \over [R]^n } K = rate , when \;n = 0 $
The rate constant of reaction is $ 3 \times 10^{-3} bar^{-1} sec^{-1} $ . The order of reaction is ...
$ 2 K = { rate \over [R]^n } = { bar /s \over bar^n } when n = 2 , k = bar ^ {-1} S^{-1} $
Which of the following statements is incorrect about the molecularity of a reaction ?
There is no difference between order and molecularity of a reaction.
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