Question

For the decomposition of azo-isopropane to hexane and nitrogen at 543K, the following data are obtained.

t(sec)	P(mm of Hg)
0	35.0
360	54.0
720	63

Calculate the rate constant

Answer 1

Chemical reactions are assigned reaction orders that describe their kinetics. A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. For a reaction $A \to$ products. This is a first-order reaction. The rate equation of the first-order reaction is $r = k\left[ A \right]$ . Where, the rate is r, the rate constant is k, and [A] is the concentration of reactant A at a time t. The unit of the rate depends upon the concentration of reactant and rate constant.

Formula used: $k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{{P_0} - P}}} \right)$

Complete step by step answer:
The rate of a reaction is the speed at which a chemical reaction takes place. If a reaction has a low rate, that means the molecules combine at a slower speed than a reaction at a high rate.
Some reactions take hundreds or maybe even thousands of years while others can happen in less than one second.
Here the reaction is as follows,

$${(C{H_3})_2}CHN = N{(C{H_3})_2} \to {N_{2}} + {C_6}{H_{14}} \\\ t = 0\;\,\,\,\,\,\,\,\,\,{P_o}\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\;\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ at\;t = {t_f}\;\;{P_o} - p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\,\,\,\,\,\,\,\,\,\,\,\,\,\,p \\\$$

Now from the rate equation of 1st order reaction,
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{{P_0} - P}}} \right)$
At $t = {t_f}$ the total pressure is, $= ({P_o} - P) + P + P \Rightarrow {P_o} + P$
Therefore,

$${P_t} = {P_o} + P \\\ P = {P_t} - {P_o} \\\$$

Put the value in the rate as follows,

$$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{{P_0} - P}}} \right) \\\ k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{2{P_0} - {P_t}}}} \right) \\\$$

Now put the values of time and pressure at a different time as follows,

$$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{2{P_0} - {P_t}}}} \right) \\\ k = \dfrac{{2.303}}{{360}}\log \left( {\dfrac{{35}}{{2 \times 35 - 54}}} \right) \\\ k = \dfrac{{2.303}}{{360}}\log \left( {\dfrac{{35}}{{70 - 54}}} \right) \\\ k = \dfrac{{2.303}}{{360}}\log \left( {\dfrac{{35}}{{16}}} \right) \\\ k = \dfrac{{2.303}}{{360}}\log \left( {\dfrac{{35}}{{16}}} \right) \\\ k = 2.175 \times {10^{ - 3}}{s^{ - 1}}\; \\\$$

And when $t = 720$ , the rate constant is,

$$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_0}}}{{2{P_0} - {P_t}}}} \right) \\\ k = \dfrac{{2.303}}{{720}}\log \left( {\dfrac{{35}}{{2 \times 35 - 63}}} \right) \\\ k = \dfrac{{2.303}}{{720}}\log \left( {\dfrac{{35}}{{70 - 63}}} \right) \\\ k = \dfrac{{2.303}}{{720}}\log \left( {\dfrac{{35}}{7}} \right) \\\ k = \dfrac{{2.303}}{{720}}\log \left( 5 \right) \\\ k = 2.235 \times {10^{ - 3}}{s^{ - 1}}\; \\\$$

**Therefore, the rate constant is,

K = \dfrac{{{K_1} + {K_2}}}{2} \\\ = \dfrac{{(2.175 \times {{10}^{ - 3}} + 2.235 \times {{10}^{ - 3}})}}{2} \\\ = 2.21 \times {10^{ - 3}}{s^{ - 1}} \\\ $$ ** **Note:** For a reversible reaction at a situation when the amount of product is formed equal to the amount of reactant is formed then it is called equilibrium. At equilibrium the amount of product and reactant becomes constant. For a reaction $$A + 2B \rightleftharpoons 2C$$ , therefore, the rates of forwarding and backward reactions are, $${R_f} = {k_f}\left[ A \right]{\left[ B \right]^2}$$ and $${\text{ }}{R_b} = {k_b}{\left[ C \right]^2}$$ respectively. Now, at equilibrium, the forward and backward reaction rates become the same. As a result,

{R_f} = {R_b} \\
or,{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\
or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
{k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\

$$$$