Question

For a second order reaction
$k=\frac{1}{t}.\frac{x}{a(a-x)}$ if the concentration of two reactants A and B are :

• A. [A] = [B]
• B. [A] > [B]
• C. [A] < [B]
• D. In all cases

Answer 1

Answer: (A)

[A] = [B]

For a second-order reaction in which both reactants are same, we have:-

$2A⟶P$

Rate of the reaction=$r=\frac{dx}{dt}=K(a-x{)}^2$ $-\left(i\right)$

where, $a$ is initial concentration of $A$

$x$= concentration of product formed after time $t$

Now, separating variables of $e{q}^n\left(i\right)$ & integrating :-

$\int \frac{dx}{(a-x{)}^2}=$$\int Kdt$

$\left[\frac{-1}{a-x}\right](-1)=Kt+C$

where, $C$ is constant of integration

$\Rightarrow \frac{1}{a-x}=Kt+C$

when, $t=0$; $x=0$ so, $C=1/a$.

Hence, $\frac{1}{a-x}=kt+\frac{1}{a}$

$\Rightarrow Kt=[\frac{1}{a-x}-\frac{1}{a}]$

$\Rightarrow K=\frac{1}{t}.[\frac{1}{a-x}-\frac{1}{a}]=\frac{1}{t}.\frac{x}{a(a-x)}$