Question

Calculate \[\Delta {G^ \circ }\] for the following reaction:
\[C{O_{(g)}} + \frac{1}{2}{O_2} \to C{O_{2\left( g \right)}}\,;\,\Delta {H^ \circ } = \] -282.84 kJ
Given: \[{S^ \circ }_{C{O_2}}\] = 213.8 \[J{K^{ - 1}}mo{l^{ - 1}}\]
\[{S^ \circ }_{CO}\] = 197.9 \[J{K^{ - 1}}mo{l^{ - 1}}\]
\[{S^ \circ }_{{O_2}}\] = 205.0 \[J{K^{ - 1}}mo{l^{ - 1}}\]

Answer 1

Hint: Gibbs free energy is a thermodynamic quantity which is equal to the enthalpy of a system or a process minus the product of the entropy and the absolute temperature. The feasibility of a reaction can be determined by the value of Gibbs free energy.

Complete step-by-step answer:
In thermodynamics, Gibbs free energy is equal to the enthalpy of a system or a process minus the product of the entropy and the absolute temperature. It is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. The Gibbs free energy is the maximum amount of non – PV work (non-expansion work) that can be extracted from a thermodynamically closed system. It is denoted by G. Mathematically, it is equal to:
\[G = H - TS\]
The change in Gibbs free energy at normal temperature and pressure is equal to
\[\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }\]… eq.1
For the complete reaction, the change in entropy is equal to the sum of entropy of the products minus the sum of entropy of the reactants.
\[\Delta {S^ \circ } = {S^ \circ }_{C{O_2}} - ({S^ \circ }_{CO} + {S^ \circ }_{{O_2}})\]
\[ \Rightarrow \Delta {S^ \circ }\] = 213.8 – (197.9 + 205.0)
\[ \Rightarrow \Delta {S^ \circ }\] = -189.1

Now, using the values in equation 1, we get
\[\Delta {G^ \circ }\] = -282840 – (298)(-189.1)
     = -226.488 kJ

Hence, the standard change in Gibbs free energy is equal to -226.488 kJ.

Note: Remember that at constant pressure and temperature, the change in Gibbs free energy determines the feasibility of a reaction. If the change in Gibbs free energy is negative, then the reaction is feasible in the forward direction.