Question

The factor of $\Delta G$ value is important in metallurgy. The $\Delta G$ value for the following reactions at ${800^ \circ }{\text{C}}$ are given as:
${{\text{S}}_{\text{2}}}\left( {\text{g}} \right) + {\text{2}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);\Delta G = - 544{\text{ kJ}}$
${\text{2Zn}}\left( {\text{s}} \right) + {{\text{S}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnS}}\left( {\text{s}} \right);\Delta G = - 293{\text{ kJ}}$
${\text{2Zn}}\left( {\text{s}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right);\Delta G = - 480{\text{ kJ}}$
The $\Delta G$ for the reaction,
${\text{2ZnS}}\left( {\text{s}} \right) + 3{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right) + 2{\text{S}}{{\text{O}}_2}\left( {\text{g}} \right)$ will be:
A) $ - 357{\text{ kJ}}$
B) $ - 731{\text{ kJ}}$
C) $ - 773{\text{ kJ}}$
D) $ - 229{\text{ kJ}}$

Answer 1

Hint: The given reaction for which $\Delta G$ is to be calculated can be obtained by rearranging the three reactions. Check for the desired reactants and products and rearrange the reactions.

Complete step-by-step answer:
The given reactions are as follows:
${{\text{S}}_{\text{2}}}\left( {\text{g}} \right) + {\text{2}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);\Delta G = - 544{\text{ kJ}}$ …… (1)
${\text{2Zn}}\left( {\text{s}} \right) + {{\text{S}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnS}}\left( {\text{s}} \right);\Delta G = - 293{\text{ kJ}}$ …… (2)
${\text{2Zn}}\left( {\text{s}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right);\Delta G = - 480{\text{ kJ}}$ …… (3)
We want to calculate $\Delta G$ for the reaction,
${\text{2ZnS}}\left( {\text{s}} \right) + 3{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right) + 2{\text{S}}{{\text{O}}_2}\left( {\text{g}} \right)$
In this reaction, ${\text{2ZnS}}\left( {\text{g}} \right)$ is on the reactant side. In reaction (2) ${\text{2ZnS}}\left( {\text{g}} \right)$ is on the product side. Thus, we reverse reaction (2) as follows:
${\text{2ZnS}}\left( {\text{s}} \right) \to {\text{2Zn}}\left( {\text{s}} \right) + {{\text{S}}_{\text{2}}}\left( {\text{g}} \right);\Delta G = + 293{\text{ kJ}}$ …… (4)
When the reaction is reversed, the negative sign of $\Delta G$ becomes positive.
The desired reaction can be obtained by adding reaction (1), (2) and (4). Thus,
${{\text{S}}_{\text{2}}}\left( {\text{g}} \right) + {\text{2}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);\Delta G = - 544{\text{ kJ}}$ …… (1)
${\text{2Zn}}\left( {\text{s}} \right) + {{\text{S}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnS}}\left( {\text{s}} \right);\Delta G = - 293{\text{ kJ}}$ …… (2)
$\underline {{\text{2ZnS}}\left( {\text{s}} \right) \to {\text{2Zn}}\left( {\text{s}} \right) + {{\text{S}}_{\text{2}}}\left( {\text{g}} \right);\Delta G = + 293{\text{ kJ}}} $ …… (4)
${\text{2ZnS}}\left( {\text{s}} \right) + 3{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right) + 2{\text{S}}{{\text{O}}_2}\left( {\text{g}} \right);\Delta G = - 731{\text{ kJ}}$
The value of $\Delta G$ for the desired reaction can be obtained by adding the $\Delta G$ values of the reactions (1), (2) and (4).
Thus the $\Delta G$ for the reaction ${\text{2ZnS}}\left( {\text{s}} \right) + 3{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{2ZnO}}\left( {\text{s}} \right) + 2{\text{S}}{{\text{O}}_2}\left( {\text{g}} \right)$ is $ - 731{\text{ kJ}}$.

Thus, the correct answer is option ‘B’ $ - 731{\text{ kJ}}$.

Note: The energy associated with any chemical reaction which can be used to perform work is known as the Gibbs free energy $\left( {\Delta G} \right)$ of the system. The free energy of the system is the summation of the enthalpy of the system and the product of temperature and entropy of the system. When the free energy is negative, a spontaneous reaction occurs. When the free energy is positive, a nonspontaneous reaction occurs. When the free energy is equal to zero, the system has achieved a state of equilibrium. Thus, the free energy of any system in nature is always negative. The change in free energy tells us the direction and the extent of the reaction.