Question

For the reaction \[{A_{\left( s \right)}} \rightleftharpoons {B_{\left( g \right)}} + {C_{\left( g \right)}}\] . What will be the value of the natural logarithm of the ratio of total pressure at \[400K\] to that at \[300K\] . \[\left[ {\ln \dfrac{{{P_{400}}}}{{{P_{300}}}}} \right]\] if \[\Delta H = 16.628kJ\] Given \[R = 8.314J{\left( {K.mole} \right)^{ - 1}}\]
A. \[\dfrac{5}{3}\]
B. \[\dfrac{5}{6}\]
C. \[\dfrac{3}{5}\]
D. \[\dfrac{6}{5}\]

Answer 1

Hint: The enthalpy is the hat content in a system. The change in enthalpy, ideal gas constant, temperature of both final and initial states and pressure of a system in both the initial state and final state were related in the Clausius-Clapeyron equation. The natural logarithm of the ratio of total pressure at \[400K\] to that at \[300K\] can be determined from the equation.
Formula used:
\[\ln \dfrac{{{P_1}}}{{{P_2}}} = - \dfrac{{\Delta H}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]\]
\[{P_1}\] is pressure at \[400K\]
\[{P_2}\] is pressure at \[300K\]
\[\Delta H\] is change in enthalpy
R is ideal gas constant
\[{T_1}\] is \[400K\]
\[{T_2}\] is \[300K\]

Complete step-by-step answer:
We have the values of temperature of final and initial. The ideal gas constant is already given as \[R = 8.314J{\left( {K.mole} \right)^{ - 1}}\] . The change in enthalpy is also given as \[\Delta H = 16.628kJ\] .
Substitute all these values in the above formula,
\[\ln \dfrac{{{P_1}}}{{{P_2}}} = - \dfrac{{16628}}{{8.314}}\left[ {\dfrac{1}{{400}} - \dfrac{1}{{300}}} \right]\]
Here, the given value of change in enthalpy is in Joules, this should be converted into joules by multiplying with \[1000\] .
\[\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = - \dfrac{{16628}}{{8.314}}\left[ {\dfrac{1}{{400}} - \dfrac{1}{{300}}} \right]\]
By simplification we will get
\[\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = - 2000 \times \dfrac{{\left( { - 100} \right)}}{{1200}}\]
Further simplification, the value will be
\[\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = \dfrac{5}{3}\]
Thus, the natural logarithm of the ratio of total pressure at \[400K\] to that at \[300K\] is \[\dfrac{5}{3}\] .

So, the correct answer is “Option A”.

Note:
If the enthalpy change is positive then the reaction favours an endothermic reaction, as less energy is released and if the enthalpy change is negative then the reaction favours an exothermic reaction. In an exothermic reaction the heat releases and the temperature increases.