Question

(a) For a reaction\[A + B \to P\], the rate is given by
Rate=\[k\left[ A \right]{\left[ B \right]^2}\]
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) 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: We must understand that “A” is present in large amounts then the reaction will be independent of A and hence it will depend on reactant B.

Complete step by step answer:
Let’s start with discussing the rate of reaction that is given to us. According to the rate of reaction given the effect that reactant B will have on rate of reaction is more as compared to the reactant A. The k is the rate constant which varies reaction to reaction so for this reaction its value will be different.
Coming to part (i)
This part states that what will happen if concentration of B is doubled. We are given the rate of reaction as
Rate =\[\;k\left[ A \right]{\left[ B \right]^2}\]
Since, the concentration of B changes to twice of B i.e. 2B we will replace B from the equation and put 2B there and solve
\[Rate{\text{ }} = {\text{ }}k\left[ A \right]{\left[ {2B} \right]^2}\]
\[Rate{\text{ }} = {\text{ }}4k\left[ A \right]{\left[ B \right]^2}\]
It is clearly visible that the rate is increased four times when the concentration of B is doubled.
Coming to (ii)
This part states that what will be the overall order if concentration of A is present in large amounts. So, the answer to this will be, since, A is present in large amounts then the reaction will be independent of A and hence it will depend on reactant B. Therefore, the overall order of the reaction will be 2.
b) For first order reaction we know that k = 0.693/t1/2 , now using the information given we will find k and then using it we will find t for 90% reaction
\[k = \dfrac{{0.693}}{{30}} = 0.0231\]
We also know that \[t = \left( {\dfrac{{2.303}}{k}} \right) \times log\left( {\dfrac{1}{1} - conc.} \right)\]
\[T = \left( {\dfrac{{2.303}}{{0.0231}}} \right) \times log\left( {\dfrac{{100}}{{10}}} \right) = 100min\]
So the time taken for 90% completion will be 100 min.

Note:
We must know that the reaction kinetics is a field of chemistry which deals with the rate at which reaction will happen along with what that rate is dependent, so that we can achieve the required benefits or effects in a stipulated time period. It also tells how much reaction is being done at a given time period.
We must know that as all the reactants increase in concentration, more molecules or ions interact to form new compounds, and the rate of reaction increases. When a reactant's concentration decreases, fewer of that molecule or ion is present, and the reaction rate decreases.