Question

Calculate the partial pressures of reactants and products, when azomethane at an initial pressures of 200mm for $30\min $ according to ${\left( {C{H_3}} \right)_2}{N_2} \to {C_2}{H_6} + {N_2}$ .
The rate constant is $2.5 \times {10^{ - 4}}{\sec ^{ - 1}}$
A. $P'{\left( {C{H_3}} \right)_2}{N_2} = 12.755mm,P'{C_2}{H_6} = 72.45mm,P'{N_2} = 724.5mm$
B. $P'{\left( {C{H_3}} \right)_2}{N_2} = 127.55mm,P'{C_2}{H_6} = 72.45mm,P'{N_2} = 72.45mm$
C. $P'{\left( {C{H_3}} \right)_2}{N_2} = 1275.5mm,P'{C_2}{H_6} = 72.45mm,P'{N_2} = 724.5mm$
D. None of these

Answer 1

Hint: Partial pressures is defined as the measure of the pressure exerted by individual components present in a gas. It is the measure of thermodynamic activity. The given reaction is a type of first order reaction.

Complete step by step answer:
${\left( {C{H_3}} \right)_2}{N_2} \to {C_2}{H_6} + {N_2}$ .
In the reaction given above, Azomethane reacts to give ethane and nitrogen gas respectively.
First order reaction: It is the overall rate of the reaction that is proportional to the first power of the concentration of the reactants only.
Rate equation for the first order reaction is given as follows:
$k = \dfrac{{2.303}}{t}\log \dfrac{A}{{\left( {A - x} \right)}}$ where, $k = $ rate constant, $t = $ time, $A = $Concentration, $A - x = $ concentration $A$ unreacted in time $'t'$ .
Similarly, we will write this equation in terms of pressure we get:
$k = \dfrac{{2.303}}{t}\log \dfrac{P}{{\left( {P - x} \right)}}$ where, $k = $ rate constant, $t = $ time, $P = $ Pressure, $P - x = $ pressure $P$ unreacted in time $'t'$ .
Using the above equation in terms of pressure we will find the partial pressures of reactants and products respectively.
First we will find the partial pressure of azomethane
Given data: Rate constant $ = 2.5 \times {10^{ - 4}}{\sec ^{ - 1}}P = 200mm$
$t = 30\min $ therefore, $t = 1800\sec $
To find: $P - x$
Formula to be used: $k = \dfrac{{2.303}}{t}\log \dfrac{P}{{\left( {P - x} \right)}}$
Substituting the values we get,
$ \Rightarrow 2.5 \times {10^{ - 4}} = \dfrac{{2.303}}{{1800}}\log \dfrac{{200}}{{\left( {P - x} \right)}}$
$ \Rightarrow 2.5 \times {10^{ - 4}} = 0.001279\log \dfrac{{200}}{{\left( {P - x} \right)}}$
On rearranging the terms, we get
$ \Rightarrow \dfrac{{2.5 \times {{10}^{ - 4}}}}{{0.001279}} = \log \dfrac{{200}}{{\left( {P - x} \right)}}$
$ \Rightarrow 0.1954 = \log 200 - \log \left( {P - x} \right)$
$ \Rightarrow 0.1954 = 2.3010 - \log \left( {P - x} \right)$
On further solving, we get
$ \Rightarrow \log \left( {P - X} \right) = 2.3010 - 0.1954$
$ \Rightarrow \log \left( {P - X} \right) = 2.106$
on taking the antilog of above value, we get
Therefore, $P - x = 127.55mmHg$
Using this partial pressure of azomethane and initial pressure we will find the partial pressure of the products respectively:
Partial pressures of products $ = $ Initial pressure $ - $ partial pressure of azomethane
Substituting the values we get,
Partial pressures of products $ = 200 - 127.55$
Partial pressures of products $ = 200 - 127.55$
Partial pressures of products $ = 72.45mmHg$
Thus the partial pressure of the reactant and products are $P'{\left( {C{H_3}} \right)_2}{N_2} = 127.55mm,P'{C_2}{H_6} = 72.45mm,P'{N_2} = 72.45mm$ respectively.

So, the correct answer is Option A.

Note: In a mixture partial pressure depends on the fraction of its mole. It is the sum of individual pressures of each component which combines to give the overall pressure of the gas. Pressure exerted by each gas is independent of each other.