Question

(i) For a reaction, A + B → P, the rate is given by Rate = k[A][B]²

(a) How is the rate of reaction affected if the concentration of B is doubled?

(b) What is the overall order of reaction if A is present in large excess?

(ii) A first order reaction takes 30 min for 50% completion. Calculate the time required for 90% completion of this reaction.

(log 2 = 0.3010)

Answer 1

(i) A + B → P

Raté = k[A][B]² (given)

(a) If concentration of B is doubled, then rate of reaction= k[A][2B]² = 4k[A][B]²

∴ Rate becomes 4 times the original rate.

(b) If A is present in large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B. As [B]² will be the only determining factor in the order of reaction, the overall order of the reaction will be two.

(ii) For the given first order reaction, the rate constant for 50% completion is given by 

k = \(\frac{2.303}{t}\ log\ \frac{[R]_0}{[R]}\)

Here, t = time taken for 50% completion = 30 min 

[R]₀ = initial concentration of reactant

[R] = final concentration of reactant

Let [R]o be 100 and due to 50% completion of reaction, [R] will be

100 - 50, i.e. 50

Putting values in (i), we get

k = \(\frac{2.303}{30}\ log\ \frac{100}{50}\) 

= \(\frac{2.303}{30}\ log\ 2\) 

 = 0.023 min¹

For same reaction, the time required for 90% comple- tion of reaction can be computed using the expression,

k = \(\frac{2.303}{30}\ log\ \frac{[R]_0}{[R]}\) 

Here, [R] = final concentration of reactant = 100 - 90 =

0.023 = \(\frac{2.303}{t}\ log\ \frac{100}{10}\)

= t = \(\frac{2.303}{0.023}\ log\ 10\)

= 100.13 min

Therefore, the time required for 90% completion of the given first order reaction is 100.13 min.