Question

A. For a reaction $A + B \to P$ , the rate is given by $Rate = \left[ A \right]{\left[ B \right]^2}$
How is the rate of reaction affected if the concentration of B is doubled?
What is the overall order of reaction if A is present in large excess?
B. A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.

Answer 1

Hint: Factors that can affect the rate of reaction are concentration, catalyst, pressure, temperature. When two reactants react with each other they come up with the product of reaction.

Complete step by step answer:
Rate of reaction may be given as the change in concentration of reactant or product to per unit time. There are several factors on which rate of reaction depends, pressure, temperature, concentration, catalyst.
Rate of reaction is directly proportional to the concentration of reactant,
$R \propto k\left[ A \right]\left[ B \right]$ ,
 Where $R = $ rate of reaction,$k = $ rate constant, $A,B$ =concentration of reactant.
Now, comes to the solution part rate of reaction given as –
1. $R = k\left[ A \right]{\left[ B \right]^2}$ Equation number 1
When concentration of B is doubled then rate of reaction may be given as –
$R = \left[ A \right]{\left[ {2B} \right]^2}$
$R = 4\left[ A \right]{\left[ B \right]^2}$ Equation number 2
Comparing equation 1 and equation 2 it is clear that the rate of reaction becomes four times.
2. When A is present in large excess, which means rate will not depend upon the concentration on A, rate of reaction will only depend upon the concentration of B. Then , the overall order of reaction will be 2 .
b. 50% completion means the item is decomposes one half of the actual concentration that is the value of half-life is given which is 30 min, then for 90% completion,
For first order reaction we have, $K = 0.693/{t_{1/2}}$
$K = 0.693/30 = 0.0231$
The time required to complete 90% reaction,
$T = 2.303\left( {\log {A_0}/A} \right)/k$
$T = $ time taken,${A_0} = \text{initial concentration}$ ,$A = \text{final concentration}$ ,$k = $ rate constant.
Then, $T = 2.303\left( {\log 100/10} \right)/.0231$
$T = 100 minutes$

Hence, the required time is 100 minutes.

Note: All radioactive reaction follows first order reaction. First order reaction does not depend upon initial concentration. Half-life of reaction is inversely proportional to rate constant.