Question

For the non - stoichiometric relation :
$2A+B\to C+D$, the following kinetic data was obtained in the seperate experiments, all at 298 K.
$\begin{array}{ccc}\text{Initial concentration}& \text{Initial concentration}& \text{Initial rate of}\\ & & \text{formation}\\ \text{[A]}& \text{[B]}& {\text{(mol L}}^{-1}{\text{s}}^{-1})\\ 0.1M& 0.1M& 1.2\times {10}^{-3}\\ 0.1M& 0.2M& 1.2\times {10}^{-3}\\ 0.2M& 0.1M& 2.4\times {10}^{-3}\end{array}$

The rate law for the formation of $C$ is:

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

Answer 1

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

For the reaction, $2A+B\to C+D$
Rate of the reaction
$=-\frac{1}{2}\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=\frac{d\left[C\right]}{dt}=\frac{d\left[D\right]}{dt}$
Now, rate of reaction, $\frac{d\left[C\right]}{dt}=k\left[A{]}^x\right[B{]}^y$
From the table,
$1.2\times {10}^{-3}=k\left(0.1{)}^x\right(0.1{)}^y\dots \dots \left(i\right)$
$1.2\times {10}^{-3}=k\left(0.1{)}^x\right(0.2{)}^y\dots \dots \left(ii\right)$
$2.4\times {10}^{-3}=k\left(0.2{)}^x\right(0.1{)}^y\dots \dots \left(iii\right)$
On dividing equation (i) by (ii), we get
$\frac{1.2\times {10}^{-3}}{1.2\times {10}^{-3}}=\frac{k\left(0.1{)}^x\right(0.1{)}^y}{\left(0.1{)}^x\right(0.2{)}^y}$
$=1={\left(\frac{1}{2}\right)}^y\Rightarrow y=0$
On dividing equation (i) by (iii), we get:
$\frac{1.2\times {10}^{-3}}{2.4\times {10}^{-3}}=\frac{k\left(0.1{)}^x\right(0.1{)}^y}{k\left(0.2{)}^x\right(0.1{)}^y}$
${\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^x\Rightarrow x=1$
Hence, $\frac{d\left[C\right]}{dt}=k\left[A{]}^1\right[B{]}^0=k\left[A\right]$