Question

Compounds ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ react according to the following chemical equation. $A_{\left(g\right)}+2B_{\left(g\right)}\to 2C_{\left(g\right)}$
Concentration of either ‘A’ or ‘B ’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment	Initial concentration of $\left[A\right]/mol{L}^{-1}$	Initial concentration of $\left[B\right]/mol{L}^{-1}$	Initial rate of formation of $\left[C\right]/mol{L}^{-1}$
1.	0.30	0.30	0.10
2.	0.30	0.60	0.40
3.	0.60	0.30	0.20

• A. Rate =$k{\left[A\right]}^2$ $\left[B\right]$
• B. Rate =$k\left[A\right]$ ${\left[B\right]}^2$
• C. Rate=$k\left[A\right]$ $\left[B\right]$
• D. Rate=$k{\left[A\right]}^2$ ${\left[B\right]}^0$

Answer 1

Answer: (B)

Let order with respect to $A$ and $B$ are $x$ and $y$ respectively.
$\therefore$ Rate$=k{\left(A\right)}^x{\left(B\right)}^y$
$0.1=k{\left(0.3\right)}^x$ ${\left(0.3\right)}^y$ $\dots \left(i\right)$
$0.4=k{\left(0.3\right)}^x$ ${\left(0.6\right)}^y$ $\left(ii\right)$
$0.2=k{\left(0.6\right)}^x$ ${\left(0.3\right)}^y$ $\left(iii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$
$\frac{0.4}{0.1}$ $=\frac{{\left(0.6\right)}^y}{{\left(0.3\right)}^y}$
Dividing $\left(iii\right)$ by $\left(i\right)$
$\frac{0.2}{0.1}$ $=\frac{{\left(0.6\right)}^x}{{\left(0.3\right)}^x}$
$\therefore$ $x=1$
Rate law will be :
Rate$=k{\left[A\right]}^1$ ${\left[B\right]}^2$