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Question: For the reaction \[{A_{\left( s \right)}} \rightleftharpoons {B_{\left( g \right)}} + {C_{\left( g \...

For the reaction A(s)⇌B(g)+C(g){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 400K400K to that at 300K300K . [ln⁡P400P300]\left[ {\ln \dfrac{{{P_{400}}}}{{{P_{300}}}}} \right] if ΔH=16.628kJ\Delta H = 16.628kJ Given R=8.314J(K.mole)−1R = 8.314J{\left( {K.mole} \right)^{ - 1}}
A. 53\dfrac{5}{3}
B. 56\dfrac{5}{6}
C. 35\dfrac{3}{5}
D. 65\dfrac{6}{5}

Explanation

Solution

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 400K400K to that at 300K300K can be determined from the equation.
Formula used:
ln⁡P1P2=−ΔHR[1T1−1T2]\ln \dfrac{{{P_1}}}{{{P_2}}} = - \dfrac{{\Delta H}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]
P1{P_1} is pressure at 400K400K
P2{P_2} is pressure at 300K300K
ΔH\Delta H is change in enthalpy
R is ideal gas constant
T1{T_1} is 400K400K
T2{T_2} is 300K300K

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(K.mole)−1R = 8.314J{\left( {K.mole} \right)^{ - 1}} . The change in enthalpy is also given as ΔH=16.628kJ\Delta H = 16.628kJ .
Substitute all these values in the above formula,
ln⁡P1P2=−166288.314[1400−1300]\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 10001000 .
ln⁡P400P300=−166288.314[1400−1300]\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = - \dfrac{{16628}}{{8.314}}\left[ {\dfrac{1}{{400}} - \dfrac{1}{{300}}} \right]
By simplification we will get
ln⁡P400P300=−2000×(−100)1200\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = - 2000 \times \dfrac{{\left( { - 100} \right)}}{{1200}}
Further simplification, the value will be
ln⁡P400P300=53\ln \dfrac{{{P_{400}}}}{{{P_{300}}}} = \dfrac{5}{3}
Thus, the natural logarithm of the ratio of total pressure at 400K400K to that at 300K300K is 53\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.