Question

Compounds 'A' and 'B' react according to the following chemical equation.

$A\left(g\right)+2B\left(g\right)⟶2C\left(g\right)$

Concentration of either 'A' or 'B' were changed keeping the concentration 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.

$\begin{array}{cccc}Experiment& InitialConcentrationof\left[A\right]/moI{L}^{-1}& Initialcooncentrationof\left[B\right]/moI{L}^{-1}& Initialcooncentrationof\left[C\right]/moI{L}^{-1}{s}^{-1}\\ 1.& 0.30& 0.30& 0.10\\ 2.& 0.30& 0.60& 0.40\\ 3.& 0.60& 0.30& 0.20\end{array}$
$\left(a\right)Rate=k\left[A{]}^2\right[B\left]\right(b)Rate=k[A\left]\right[B{]}^2\left(c\right)Rate=k\left[A\right]\left[B\right]\left(d\right)Rate=k\left[A{]}^2\right[B{]}^0$

Answer 1

Rate of reaction change in concentration of reactant with respect to time.

$r=k\left[A{]}^x\right[B{]}^y\frac{Rateofexp.1}{Rateofexp.2}=\frac{\left[0.30{]}^x\right[0.30{]}^y}{\left[0.30{]}^x\right[0.60{]}^y}\frac{0.10}{0.40}=\frac{[0.30{]}^y}{[0.60{]}^y}\frac{1}{4}={\left[\frac{1}{2}\right]}^y{\left[\frac{1}{2}\right]}^2={\left[\frac{1}{2}\right]}^2\frac{Rateofexp.1}{Rateofexp.3}=\frac{\left[0.30{]}^x\right[0.30{]}^y}{\left[0.60{]}^x\right[0.30{]}^y}\frac{0.10}{0.20}={\left[\frac{0.30}{0.60}\right]}^x{\left[\frac{0.30}{0.30}\right]}^y\frac{1}{2}={\left[\frac{1}{2}\right]}^x[1{]}^y\frac{1}{2}={\left[\frac{1}{2}\right]}^{x}i.e.,x=1\therefore Rate=k[A{]}^x[B{]}^{y}Rate=k[A{]}^1[B{]}^2$