Question

When the rate is determined by the change in concentration of two different reactants, then the kinetic equation may be expressed as:

• A. $k_2=\frac{2.303}{(a-b)t}log\frac{(a-x)b}{(b-x)a}$
• B. $k_2=\frac{2.303}{(a-b)t}log\frac{(a-x)}{(b+x)}$
• C. $k_2=\frac{2.303}{(a-b)}log\frac{(a-x)}{(b-x)}$
• D. $k_2=\frac{1}{t}\times \frac{x}{(a-x)}$

Answer 1

The correct option is A $k_2=\frac{2.303}{(a-b)t}log\frac{(a-x)b}{(b-x)a}$

SECOND ORDER REACTION

	A	+	A	$\to$	Product
	A	+	B	$\to$	Product
At $t=0$	a		a		$0$
At $t=t$	$(a-x)$		$(a-x)$		$x$
At $t=t_1$	$(a-x_1)$		$(a-x_1)$		$x_1$
At $t=t_2$	$(a-x_2)$		$(a-x_2)$		$x_2$

As per rate law, $\frac{dx}{dt}=k_2[A{]}^n=k_2[A{]}^2=k_2\left[A\right]\left[B\right]$

$\therefore \left(\frac{dx}{dt}\right)=k_2(a-x{)}^2$ ($k_2$= rate constant for second order reaction)

Also, $k_2=\frac{1}{t}[\frac{1}{(a-x)}-\frac{1}{a}]=\frac{1}{t}\frac{a}{a(a-x)}$ or $k_2=\frac{1}{(t_2-t_1)}[\frac{1}{(a-x_2)}-\frac{1}{(a-x_1)}]$

Where $(a-x_1)$ and $(a-x_2)$ are the concentration of the reactant A at time $t_1$ and $t_2$ respectively. If reactant $A$ and $B$ have different concentrations $a$ and $b$, then $k_2=\frac{2.3030}{t(a-b)}{\log }_{10}\frac{b(a-x)}{a(b-x)}$