Question

Consider the kinetic data given in the following table for the reaction A + B + C → Product

Ex. No.	[A]	[B]	[C]	Rate of reaction
1.	$0.2$	$0.1$	$0.1$	$6\times {10}^{-5}$
2.	$0.2$	$0.2$	$0.1$	$6\times {10}^{-5}$
3.	$0.2$	$0.1$	$0.2$	$1.2\times {10}^{-4}$
4.	$0.3$	$0.1$	$0.1$	$9\times {10}^{-5}$

When [A] $=0.15$

[B] $=0.25$

[C] $=0.15$

Rate of reaction is $Y\times {10}^{-5}m/s$. Find $\mathrm{Y}$.

Answer 1

The above question can be solved as:

Step 1:

Let us write the rate of reaction equation as: $\mathrm{r}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{a}}{\left[\mathrm{B}\right]}^{\mathrm{b}}{\left[\mathrm{C}\right]}^{\mathrm{c}}------\left(\mathrm{i}\right)$

where $\mathrm{x},\mathrm{y},\mathrm{z}$ are coefficient of reactant $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ respectively.

and, order of the reaction $=\mathrm{x}+\mathrm{y}+\mathrm{z}-------\left(\mathrm{ii}\right)$

Step 2:

Apply the data of exp($1$) in equation($\mathrm{i}$):

$6\times {10}^{-5}=\mathrm{k}{\left[0.2\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}{\left[0.1\right]}^{\mathrm{z}}-------\left(\mathrm{iii}\right)$

Again apply the date of exp($2$) in equation($\mathrm{i}$):

$6\times {10}^{-5}=\mathrm{k}{\left[0.2\right]}^{\mathrm{x}}{\left[0.2\right]}^{\mathrm{y}}{\left[0.1\right]}^{\mathrm{z}}-------\left(\mathrm{iv}\right)$

Divide eq($\mathrm{iii}$) by eq($\mathrm{iv}$)

$$\begin{gathered} \frac{6\times {10}^{-5}}{6\times {10}^{-5}}=\frac{k{\left[0.2\right]}^x{\left[0.1\right]}^y\left[0.1\right]z}{k{\left[0.2\right]}^x{\left[0.2\right]}^y{\left[0.1\right]}^z} \\ 1={\left(0.5\right)}^y \\ y=0 \end{gathered}$$

Similarly $\mathrm{x},\mathrm{z}$ can be found as $1,1$

and the order of the reaction is $$\begin{gathered} =1+0+1 \\ =2 \end{gathered}$$

So, the rate law for reaction is: $$\begin{gathered} r=k{\left[A\right]}^1{\left[B\right]}^0{\left[C\right]}^1 \\ r=k\left[A\right]\left[C\right]\_\_\_\_\_\_\left(v\right) \end{gathered}$$

Step 3:

Placing the values of concentration of A and C in equation $\left(v\right)$ to find the value of $\mathrm{k}$.

From experiment $2.$:

$$\begin{gathered} 6\times {10}^{-5}=k\left(0.2\times 0.1\right) \\ k=3\times {10}^{-3}mo{l}^{-1}L{s}^{-1} \end{gathered}$$

Step 4:

Now, place the given concentrations of A, B, and C and calculate the value of $\mathrm{k}$ in the equation $\left(v\right)$ to determine the rate of reaction.

$$\begin{gathered} r=3\times {10}^{-3}\times 0.15\times 0.15 \\ r=6.75\times {10}^{-5}mo{l}^{-1}L{s}^{-1} \end{gathered}$$

Therefore, as the rate of reaction is $\mathit{Y}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ then $\mathit{Y}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{75}$.