Question

Compounds $'A'$ and $'B'$ react according to the following chemical equation. $A_{\left(g\right)}+2B_{\left(g\right)} \rightarrow2C_{\left(g\right)}$
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 $ $\quad$ Rate$=k\left(A\right)^{x} \left(B\right)^{y}$
$0.1=k\left(0.3\right)^{x}$ $\left(0.3\right)^{y}$ $\quad$ $\ldots\left(i\right)$
$0.4=k\left(0.3\right)^{x}$ $\left(0.6\right)^{y}$ $\quad$ $\left(ii\right)$
$0.2=k\left(0.6\right)^{x}$ $\left(0.3\right)^{y}$ $\quad$ $\left(iii\right)$
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 $ $\quad$ $x=1$
Rate law will be :
Rate$=k\left[A\right]^{1}$ $\left[B\right]^{2}$