Question

For the non – stoichimetric reaction, $2A+B\to C+D,$ the following kinetic data were obtained in three separate experiments, all at 298 K.
$\begin{array}{ccc}InitialConcentration& InitialConcentration& InitialrateofFormationofC\\ \left(A\right)& \left(B\right)& \left(mol{L}^{-1}{s}^{-1}\right)\\ 0.1M& 0.1M& 1.2\times {10}^{-3}\\ 0.1M& 0.2M& 1.2\times {10}^{-3}\\ 0.2M& 0.1M& 2.4\times {10}^{-3}\end{array}$
The rate law for the formation of C is

• A. $\frac{dc}{dt}=k\left[A\right]\left[B\right]$
  
• B. $\frac{dc}{dt}=k\left[A{]}^2\right[B]$
  
• C. $\frac{dc}{dt}=k\left[A\right][B{]}^2$
  
• D. $\frac{dc}{dt}=k\left[A\right]$

Answer 1

The correct option is D $\frac{dc}{dt}=k\left[A\right]$

Rate law equation is given by
$\frac{dc}{dt}=k\left[A{]}^x\right[B{]}^y$
$\therefore \frac{dc}{dt}$ for C at various concentrations are
$1.2\times {10}^{-3}=k(0.1{)}^x\times (0.1{)}^y.....\left(i\right)1.2\times {10}^{-3}=k(0.1{)}^x\times (0.2{)}^y.....\left(ii\right)2.4\times {10}^{-3}=k(0.2{)}^x\times (0.1{)}^y.....\left(iii\right)$
Solving for x [divided Eq. (iii) by Eq. (i)]
$\frac{2.4\times {10}^{-3}}{1.2\times {10}^{-3}}=\frac{k\left(0.2{)}^x\right(0.1{)}^y}{k\left(0.1{)}^x\right(0.1{)}^y}2=(2{)}^x\Rightarrow x=1$
Solving for y [divide Eq. (iii) by Eq. (i)]
$\frac{2.4\times {10}^{-3}}{1.2\times {10}^{-3}}=\frac{k\left(0.2{)}^x\right(0.1{)}^y}{k\left(0.1{)}^x\right(0.2{)}^y}$
$2=(2{)}^x{\left(\frac{0.1}{0.2}\right)}^{y}1=(0.5{)}^y(1{)}^{\circ }=(0.5{)}^y\Rightarrow y=0Thus,\frac{dc}{dt}=k\left[A{]}^1\right[B{]}^0\Rightarrow \frac{dc}{dt}=k[A{]}^1$