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For the reaction $ CH_3COCH_3 + I_2 + H^+ --> Products $ , the rate is governed by, $ rate = K[CH_3COCH_3] [H^+]$ . The rate order of iodine is = _.
$ OO \therefore No I_2 $ in the rate law equation.
If the order of reaction is zero. It means that
rate of zero order reaction is independent of the concentration of the reacting species
The reactions of higher order are rare because
many body collisions have a low probability
$ 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 $
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