Question

Solid $Ca{{\left( HC{{O}_{3}} \right)}_{2}}$ decomposes as
\[Ca{{\left( HC{{O}_{3}} \right)}_{2}}\left( s \right)\rightleftharpoons CaC{{O}_{3}}\left( s \right)+C{{O}_{2}}\left( g \right)+{{H}_{2}}O\left( g \right)\]
If the total pressure is 0.2 bar at 420 K, what is the standard free energy change for the given reaction$\left( {{\Delta }_{r}}{{G}^{o}} \right)$?
(A) 840 $KJ/mol$
(B) 3.86 $KJ/mol$
(C) 6.98 $KJ/mol$
(D) 16.083 $KJ/mol$

Answer 1

Hint: When calcium bicarbonate is subjected to heating it is decomposed to give calcium carbonate, carbon dioxide and water. The calcium bicarbonate is soluble in water whereas calcium carbonate is insoluble in water.

Complete step by step solution:
- By looking at the stoichiometry of the reaction of decomposition of $Ca{{\left( HC{{O}_{3}} \right)}_{2}}$, we can assume that the partial pressure of carbon dioxide is equal to the partial pressure of water. And hence we can write as follows
\[{{P}_{C{{O}_{2}}}}={{P}_{{{H}_{2}}O}}\]
 It’s given that the total pressure is 0.2 bar. Thus we can write as follows
\[{{P}_{C{{O}_{2}}}}+{{P}_{{{H}_{2}}O}}=0.2 bar\]
We can modify the above relation as follows
\[2{{P}_{C{{O}_{2}}}}=2{{P}_{{{H}_{2}}O}}=0.2 bar\]
At equilibrium the relation becomes,
\[{{P}_{C{{O}_{2}}}}={{P}_{{{H}_{2}}O}}=0.1 bar\]
\[{{K}_{eq}}={{P}_{C{{O}_{2}}}}\times {{P}_{{{H}_{2}}O}}=0.1 ba{{r}^{2}}\]
\[{{K}_{eq}}=0.01 ba{{r}^{2}}\]
We are asked to find the ${{\Delta }_{r}}{{G}^{o}}$ value. The relation between Gibbs energy and standard free energy change can be written as follows
\[\Delta G=\Delta {{G}^{o}}+RT\ln {{K}_{eq}}\]
At equilibrium $\Delta G$ is zero and the above relation can be modified as follows
\[\Delta {{G}^{o}}=-RT\ln {{K}_{eq}}\]
As we know R is the universal gas constant and its value is $8.314Jmo{{l}^{-1}}{{K}^{-1}}$. T is the temperature in kelvin and the value is given in the question as 420 K. We have found the value of ${{K}_{eq}}$ as 0.01. Let’s substitute this values in the above equation
\[\Delta {{G}^{o}}=-8.314\times 420\times \ln (0.01)=16081.6688Jmo{{l}^{-1}}\]
We know that $1J=1{{0}^{-3}}KJ$. Thus by converting the above values in joules into kilojoules we get a follows
${{\Delta }_{r}}{{G}^{o}}=16.082 KJ/mol$
Thus the standard free energy change for the given reaction ${{\Delta }_{r}}{{G}^{o}}$ is 16.083 $KJ/mol$.

Therefore the answer is option (D) 16.083 $KJ/mol$

Note: The standard-state free energy of formation can be defined as the variation in free energy which happens when a compound is formed from its elements at standard-state conditions in their most thermodynamically stable states. It can also be defined as the difference between the free energies of constituent elements and the free energy of its substance at standard-state condition.