# For the reaction $2A+B \rightarrow C$, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is : $[A](mol\; L^{-1})$ $[B](mol\; L^{-1})$ $\\Initial Rate\\(mol\; L^{-1}s^{-1})$ 0.05 0.05 0.045 0.10 0.05 0.090 0.20 0.10 0.72 Option 1) $Rate=k[A]^{2}[B]^{2}$ Option 2) $Rate=k[A]^{2}[B]$ Option 3) $Rate=k[A][B]^{2}$   Option 4) $Rate=k[A][B]$

Rate law$\Rightarrow R=k[A]^{x}[B]^{y}$

 $[A](mol\; L^{-1})$ $[B](mol\; L^{-1})$ $\\Initial Rate\\(mol\; L^{-1}s^{-1})$ 0.05 0.05 0.045 $R_{1}$ 0.10 0.05 0.090 $R_{2}$ 0.20 0.10 0.72 $R_{3}$

$R_{1}=0.045=k[0.05]^{x}[0.05]^{y}$

$R_{2}=0.09=k[0.01]^{x}[0.05]^{y}$

$\frac{R_{2}}{R_{1}}=2=2^{x}\; \; \; \; x=1$               $\frac{R_{3}}{R_{2}}=\frac{0.72}{0.09}=\frac{k[0.2]^{x}}{k[0.1]^{x}}\frac{k[0.1]^{x}}{k[0.05]^{x}}$

$y=2$

$\therefore Rate \; law:-R=k[A][B]^{2}$

Option 1)

$Rate=k[A]^{2}[B]^{2}$

Option 2)

$Rate=k[A]^{2}[B]$

Option 3)

$Rate=k[A][B]^{2}$

Option 4)

$Rate=k[A][B]$

Exams
Articles
Questions