Question

The kinetic data for the given reaction $A\left(g\right)+2B\left(g\right)\stackrel{k}{\to }C\left(g\right)$ is provided in the following table for three experiments at 300 K

Ex. No.	[A/M]	[B/M]	Initial rate (M $se{c}^{-1}$)
1.	0.001	0.001	$6.930\times {10}^{-6}$
2.	0.02	0.01	$1.386\times {10}^{-5}$
3.	0.02	0.02	$1.386\times {10}^{-5}$

In another experiment starting with initial concentration of 0.5 and 1 M respectively for A and B at 300 K. find the rate of reaction after 50 minutes from start of experiment (in M/sec) :

• A. $6.93\times {10}^{-4}$
• B. $0.25\times {10}^{-7}$
• C. $4.33\times {10}^{-5}$
• D. $3.46\times {10}^{-4}$

Answer 1

The correct option is C $4.33\times {10}^{-5}$

According to given table from experiment $2$ to $3$ when concentration $B$ double. It does not affect rate So, rate is independent on concentration of $B$.

From experiment $1$ to $2$ when $A$ increase $=\frac{0.02}{0.01}=12times$ Rate also

increase $\frac{1.386\times {10}^{-5}}{6.93\times {10}^{-6}}2times$

So, Rate equation can be written as

$-\frac{dR}{dT}=K\left[A\right]$

$6.93\times {10}^{-6}=K\times 0.01$

$K=6.93\times {10}^{-4}$

$\left[A\right]=\left[A_0\right]e^{kt}$

Now $\left[A_0\right]=0.5$

After so minute $\left[K\right]$

$\left[A\right]=0.5e^{-6.9\times {10}^{-4}50\times 60}$

$\left[A\right]={0.5}^{-2.079}$

$=6.25\times {10}^{-2}$

$Rate=\frac{dR}{dt}=6.93\times {10}^{-4}\times 6.25\times {10}^{-2}$

$=4.33\times {10}^{-5}$