Solveeit Logo
Login

Question

Question: : Determine \(\Delta {{\text{G}}^ \circ }\) for the following reaction: \({\text{CO}}\left( {\tex...

: Determine ΔG∘\Delta {{\text{G}}^ \circ } for the following reaction:
CO(g)+12O2(g)→CO2(g); ΔH∘=−282.84 kJ{\text{CO}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);{\text{ }}\Delta {{\text{H}}^ \circ } = - 282.84{\text{ kJ}}
[Given: SCO2∘=213.8 J K−1 mol−1{\text{S}}_{{\text{C}}{{\text{O}}_2}}^ \circ = 213.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}, SCO∘=197.9 J K−1 mol−1{\text{S}}_{{\text{CO}}}^ \circ = 197.9{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}, SO2∘=205.8 J K−1 mol−1{\text{S}}_{{{\text{O}}_2}}^ \circ = 205.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}]
(1)−157.33 kJ - 157.33{\text{ kJ}}
(2)+201.033 kJ + 201.033{\text{ kJ}}
(3)−256.91 kJ - 256.91{\text{ kJ}}
(4)+257.033 kJ + 257.033{\text{ kJ}}

Explanation

Solution

The measure of randomness or disordered distribution is known as entropy. The randomness is always higher in a gaseous state. More the number of gaseous molecules higher is the entropy. To solve this we must know the expression that gives the relation between free energy, entropy and enthalpy.
Formula Used: ΔSreaction∘=ΔSproducts−ΔSreactants\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}
ΔG∘=ΔH∘−TΔS∘\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }

Complete step-by-step answer: VWe are given the reaction as follows:
CO(g)+12O2(g)→CO2(g){\text{CO}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)
We know that the measure of randomness or disordered distribution is known as entropy.
We will first calculate the change in entropy of the reaction using the equation as follows:
ΔSreaction∘=ΔSproducts−ΔSreactants\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}
Where, ΔSreaction∘\Delta {\text{S}}_{{\text{reaction}}}^ \circ is the standard change in entropy of the reaction.
We are given the values of standard entropies as SCO2∘=213.8 J K−1 mol−1{\text{S}}_{{\text{C}}{{\text{O}}_2}}^ \circ = 213.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}, SCO∘=197.9 J K−1 mol−1{\text{S}}_{{\text{CO}}}^ \circ = 197.9{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}, SO2∘=205.8 J K−1 mol−1{\text{S}}_{{{\text{O}}_2}}^ \circ = 205.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}. Thus,
ΔSreaction∘=213.8−(197.9+12×205.8)\Delta {\text{S}}_{{\text{reaction}}}^ \circ = {\text{213}}{\text{.8}} - \left( {197.9 + \dfrac{1}{2} \times 205.8} \right)
ΔSreaction∘=−87 J K−1 mol−1\Delta {\text{S}}_{{\text{reaction}}}^ \circ = - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}
Thus, the standard change in entropy of the reaction is −87 J K−1 mol−1=−87×10−3 kJ K−1 mol−1 - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} = - 87 \times {10^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}.
We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
ΔG∘=ΔH∘−TΔS∘\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }
Where, ΔG∘\Delta {{\text{G}}^ \circ } is the standard change in Gibb’s free energy,
ΔH∘\Delta {{\text{H}}^ \circ } is the standard change in enthalpy,
T{\text{T}} is the temperature,
ΔS∘\Delta {{\text{S}}^ \circ } is the standard change in entropy.
We are given the standard values. At standard condition, T=25∘C=298 K{\text{T}} = {25^ \circ }{\text{C}} = 298{\text{ K}}. We are given that ΔH∘=−282.84 kJ\Delta {{\text{H}}^ \circ } = - 282.84{\text{ kJ}}. Thus,
ΔG∘=−282.84 kJ−[298 K(−87×10−3 kJ K−1 mol−1)]\Delta {{\text{G}}^ \circ } = - 282.84{\text{ kJ}} - \left[ {{\text{298 K}}\left( { - 87 \times {{10}^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)} \right]
ΔG∘=−256.914 kJ\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}}
Thus, ΔG∘\Delta {{\text{G}}^ \circ } for the given reaction is −256.91 kJ - 256.91{\text{ kJ}}.
Thus, the correct option is (3) −256.91 kJ - 256.91{\text{ kJ}}.

Note: : We know that if the free energy change is negative then the reaction is spontaneous. If the free energy change is positive then the reaction is non-spontaneous. Here, ΔG∘=−256.914 kJ\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}} which is a negative value. Thus, the given reaction is a spontaneous reaction.