Question

Consider the following reaction:

2 ClO₂(aq) + 2 OH^-(aq) ClO₃^-(aq) + ClO₃^-(aq) + H₂O(l)

(a) The rate law for this reaction is second order in ClO₂(aq) and first order in OH^-(aq). What is the rate law for this reaction?

Rate = k [ClO₂(aq)] [OH^-(aq)]

Rate = k [ClO₂(aq)]² [OH^-(aq)]    

Rate = k [ClO₂(aq)] [OH^-(aq)]²

Rate = k [ClO₂(aq)]² [OH^-(aq)]²

Rate = k [ClO₂(aq)] [OH^-(aq)]³

Rate = k [ClO₂(aq)]⁴ [OH^-(aq)]

(b) If the rate constant for this reaction at a certain temperature is 362, what is the reaction rate when [ClO₂(aq)] = 0.0339 M and [OH^-(aq)] = 0.0786 M?

Rate = ________  M/s.



(c) What is the reaction rate when the concentration of ClO₂(aq) is doubled, to 0.0678 M while the concentration of OH^-(aq) is 0.0786 M?

Rate = ___________ M/s

Answer 1

Ans 2 :

a) Rate = k [ClO₂(aq)]² [OH^-(aq)]

The rate law for the reaction is given as the product of concentration of reactant species each multiplied with their respective orders in the reaction.

b)

Putting the values mentioned here in the rate law expression :

Rate = 362 (0.0339)²(0.0786)

Rate = 0.0327 M/s

c)

Since the order of ClO2 (aq) is 2 , so doubling the concentration of ClO2 (aq) while keeping the concentration of OH- (aq) constant will quadruple the rate of reaction.

So here , the rate of reaction = 4 x 0.0327 M/s

= 0.131 M/s