Question

The decomposition of hydrogen peroxide was studied at a particular temperature. The following data were obtained, where $\mathbf{Rate}\mathbf{=}\frac{\mathbf{-}\mathbf{d}\left[H_2{O}_2\right]}{\mathbf{dt}}$

Time (s)	$\left[H_2{O}_2\right]\left(\mathrm{mol}/L\right)$
0	1.00
$\mathbf{120}\mathbf{\pm }\mathbf{1}$	0.91
$\mathbf{300}\mathbf{\pm }\mathbf{1}$	0.78
$\mathbf{600}\mathbf{\pm }\mathbf{1}$	0.59
$\mathbf{1200}\mathbf{\pm }\mathbf{1}$	0.37
$\mathbf{1800}\mathbf{\pm }\mathbf{1}$	0.22
$\mathbf{2400}\mathbf{\pm }\mathbf{1}$	0.13
$\mathbf{3000}\mathbf{\pm }\mathbf{1}$	0.082
$\mathbf{3600}\mathbf{\pm }\mathbf{1}$	0.050

Determine the integrated state law, the differential state law, and the value of the rate constant. Calculate the${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$at4000.5after the start of the reaction.

Short answer

Decomposition of hydrogen peroxide:-

A decomposition reaction occurs when one reactant breaks down into two or more products.

Example:-

${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2$

Step-by-step solution

To decide the Rate for the reaction

To decide whether the reaction is first order or second order, we must see whether the plot of $\ln \left[{\mathrm{H}}_2{\mathrm{O}}_2\right]$versus time is a straight line for first order or the plot of $\frac{1}{{\mathrm{H}}_2{\mathrm{O}}_2}$versus time is a straight line for the second order.

Time	$\ln \left[{\mathrm{H}}_2{\mathrm{O}}_2\right]$
0	0
$120\pm 1$	-0.094
$300\pm 1$	-0.24
$600\pm 1$	-0.52
$1200\pm 1$	-0.99
$1800\pm 1$	-1.51
$2400\pm 1$	-2.04
$3000\pm 1$	-2.50
$3600\pm 1$	-2.95

Hence the plot of$\ln \left[{\mathrm{H}}_2{\mathrm{O}}_2\right]$versus time is a straight line therefore reaction will follow first order kinetics.

So, the rate law

$\mathrm{Rate}={\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}^1$

Integrated rate law.

$\frac{dx}{\left(a-x\right)}=K_{1}dt$

On integrating

$$\begin{gathered} \int \frac{\mathrm{dx}}{\left(\mathrm{a}-\mathrm{x}\right)}={\mathrm{K}}_1\int \mathrm{dt} \\ -\ln \left(\mathrm{a}-\mathrm{x}\right)={\mathrm{K}}_1\mathrm{t}+\mathrm{C}.......\left(1\right) \end{gathered}$$

If $t=0$, then $x=0$

So,

$\ln \left(a-0\right)=K_1\times 0+C$

$C=-\ln a$

Put the value C of in equation (1)

$\begin{array}{rcl}-\ln \left(a-x\right)& =& K_{1}t+\ln a\\ K_{1}t& =& \ln a-\ln \left(a-x\right)\\ K_{1}t& =& \ln \frac{a}{\left(a-x\right)}\\ K_{1}t& =& \ln \frac{{\left[A\right]}_t}{{\left[A\right]}_{\circ }}\end{array}$

OR

$\ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\mathrm{t}}={\mathrm{K}}_1\mathrm{t}+\ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\circ }........\left(2\right)$

and equation (2) is the integrated rate law for the first-order reaction.

The value of the rate constant.

We determine the rate constant K by determining the slope of $\ln \left[H_2{O}_2\right]$versus time plot $\left(\mathrm{slope}=-\mathrm{K}\right)$.

$$\begin{gathered} \mathrm{slope}=-\mathrm{K}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta x}}=\frac{0-\left(-0.094\right)}{-120} \\ =\frac{0.094}{-120} \\ =7.8\times {10}^{-4} \end{gathered}$$

Calculate the [H2O2] at 4000.5 after the start of the reaction.

To determine $\left[H_2{O}_2\right]$at 4000.5, use the integrated rate law where at $t=0,\left[H_2{O}_2\right]=1.00M$

$\begin{array}{rcl}\ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\mathrm{t}}-\ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\circ }& =& -\mathrm{Kt}\\ \ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\mathrm{t}}& =& \ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\circ }-\mathrm{Kt}\\ \ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\mathrm{t}}& =& \ln {\left[1.00\right]}_{\circ }-\left(7.8\times {10}^{-4}\right)4000.5\\ \ln {\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}_{\mathrm{t}}& =& 0+7.8\times {10}^{-4}\times 4000\\ \left[{\mathrm{H}}_2{\mathrm{O}}_2\right]& =& {\mathrm{e}}^{-3.12}\\ \left[{\mathrm{H}}_2{\mathrm{O}}_2\right]& =& -0.401\mathrm{mol}/\mathrm{L}\end{array}$