Question

$ MC{{O}_{3(s)}}\rightleftharpoons M{{O}_{(s)}}+C{{O}_{2(g)}} $
The equilibrium pressure of $ C{{O}_{2}} $ at $ 127{}^\circ C $ and $ 147{}^\circ C $ are $ 0.14 $ bar and $ 0.35 $ bar respectively. The enthalpy of the reaction is:
(A) $ 64.0\text{ }kJ/mol $
(B) $ 35.8\text{ }kJ/mol $
(C) $ 31.1\text{ }kJ/mol $
(D) $ 79.8\text{ }kJ/mol $

Answer 1

Hint :We know that the enthalpy of the combustion is defined as the heat energy which has been given out when one mole of the given substance gets burnt completely in oxygen. The combustion is a process or the reaction which is always exothermic in form.

Complete Step By Step Answer:
The enthalpy change for the reaction is always negative. So by convention we can conclude that the molar heat of the combustion given in the table is a positive value. The state, in which the rate of the forward reaction is the same as the rate of the backward reaction, is called chemical equilibrium. In chemical equilibrium, there is no net change of reactant and product concentration. The other name of chemical equilibrium is dynamic equilibrium. The Hess law states that the change in the overall enthalpy of the reaction is independent of the taken route. This law tends to follow the first law of thermodynamics. The application of it lies in calculating the change in the enthalpy values which are very difficult in obtaining them experimentally. This law is the consequence of the fact that the enthalpy is a function of state or a state function. Here the given reaction is; $ MC{{O}_{3(s)}}\rightleftharpoons M{{O}_{(s)}}+C{{O}_{2(g)}} $
Since we got; $ {{K}_{eq}}={{P}_{C{{O}_{2}}}} $ ; $ R=8.314 $ ; $ {{T}_{1}}=127{}^\circ C=400K $ and $ {{T}_{2}}=147{}^\circ C=420K $ also $ {{K}_{1}}=0.14 $ and $ {{K}_{2}}=0.35 $ .
Thus, $ \ln \left( \dfrac{{{K}_{2}}}{{{K}_{1}}} \right)=\dfrac{-\Delta H}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right) $ on substituting the values we get; $ \ln \left( \dfrac{0.35}{0.14} \right)=\dfrac{-\Delta H}{8.314}\left( \dfrac{1}{420}-\dfrac{1}{400} \right) $
Further solving we get;
 $ 0.916=\dfrac{-\Delta H}{8.314}\left( -0.000119 \right) $
From here we have to determine the value of $ \Delta H $ ;
 $ \Delta H=\dfrac{0.916\times 8.314}{0.000119} $
 $ \Rightarrow \Delta H=63996.84~~J $ or $ 63.99684~~J\approx 64.0~~kJ/mol $
Therefore, the correct answer is option A.

Note :
Remember that when you are using reaction to solve this type of problem, we try to solve our problem by reversing reaction this changes the enthalpy like in above reaction, by performing additions or subtractions of reaction this can be very confusing, to easily solve that you should try numbering your reactions and use them accordingly.