Question

A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Answer 1

We study the rate of reaction, factors affecting these rates and the mechanism by which the reaction occurs in a branch of science called chemical kinetics.

Complete step by step answer:
Rate of reaction may be defined as the change in concentration of a species (Reactant or product) per unit time.
Average Rate=(Change in concentration of species)/(Change in time)
Let us consider a reaction
$aA + bB \to$Product.
Where A and B reacts and a and b are the number of molecules of reactants A and B respectively.
By Rate Law Rate of reaction
$R\alpha {[A]^x}{[B]^y}$
$R = k{[A]^x}{[B]^y}$
Where K is constant and $x$ and $y$ are small integers. For reaction they can be negative zero or fractions.
The exponent $x$ and $y$ in the rate law indicates effectiveness of concentration change.
If $x = 1$and$y = 1$
$R = k\left[ A \right]\left[ B \right]$
The rate depends on concentration of A and B.
If concentration is doubled the rate of reaction also gets doubled.
$x$ and $y$ are the number of molecules at which rate of reaction depends.
If $x = 1,y = 1$
$x + y = 1 + 1$
$\Rightarrow x + y = 2.$
It means rate depends on one molecule of A and one molecule of B. The overall reaction is second order reaction.
Example: ${H_{2(g)}} + {I_{2(g)}} \to 2H{I_{(g)}}$
Rate$= k[{H_2}][{I_2}]$
First order in ${H_2}$and first order in ${I_2}$. So it is $1 + 1 = 2$i.e., second order reaction.
For a second order reaction
$R = k[A][B]$ . . . . . (1)
Let initial concentration be$x.$
$\therefore R = k \times x \times x$
$\Rightarrow R = {x^2}k$
If concentration becomes double $\left[ A \right] = 2x{\text{ }}\left[ B \right] = 2x$
Putting this value in equation (1)
$R' = k[2x][2x]$
$\Rightarrow R' = 4{x^2}k$
$R'$ is the new rate of reaction when the concentration is doubled.
$\Rightarrow\dfrac{{R'}}{R} = \dfrac{{4{x^2}k}}{{{x^2}k}}$
$\Rightarrow\dfrac{{R'}}{R} = 2$
$R' = 4R$
Therefore, rate of reaction becomes four times. If concentration becomes half i.e., $[A] = \left[ {\dfrac{x}{2}} \right]$ $[B] = \left[ {\dfrac{x}{2}} \right]$ Let us say that the rate of reaction is $R''$
Putting this value in equation (1), we get
$R = k\left[ {\dfrac{x}{2}} \right]\left[ {\dfrac{x}{2}} \right]$
The Rate becomes ${\dfrac{1}{4}^{th}}$times.
Note: The order of a reaction is not related to the stoichiometric equation of the reaction.
It is experimentally determined. It is important to note that the reaction order is always defined in terms of concentration of reactants and not to the product.