Question

If dx/dt = k[H+]n and rate becomes 100 times when pH changes from 2 to 1 the order of reaction is

Answer 1

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Answer 2

Answer: Order of the reaction is 2.

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=\frac{dx}{dt}=k[H^+]^n$

k= rate constant

n = order  

$pH=-log[H^+]$

when pH=2, $[H^+]=10^{-2}$

pH=1, $[H^+]=10^{-1}$

a) From trial 1

$Rate=k[10^{-2}]^n$    (1)

b) From trial 2: $100\times Rate=k[10^{-1}]^n$    (2)

Dividing 2 by 1 :$\frac{100\times Rate}{Rate}=\frac{k[10^{-1}]^n}{k[10^{-2}]^n}$

$100=10^n$

$10^2=10^n$

$n=2$