From these two reactions at 298 K,
V2O3(s) + 3CO(g) -> 2V(s) + 3CO2(g); DH° = 369.8 kJ; DS° = 8.3 J/K
V2O5(s) + 2CO(g) -> V2O3(s) + 2CO2(g); DH° = –234.2 kJ; DS° = 0.2 J/K
calculate DG° for the following at 298
K:
2V(s) + 5CO2(g) -> V2O5(s) + 5CO(g)
133.1kJ
To calculate the standard Gibbs free energy change (ΔG°) for a reaction, you can use the equation:
ΔG° = ΔH° - TΔS°
where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.
Given the following reactions and their respective values at 298 K:
Reaction 1:
V2O3(s) + 3CO(g) -> 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K
Reaction 2:
V2O5(s) + 2CO(g) -> V2O3(s) + 2CO2(g); ΔH° = –234.2 kJ; ΔS° = 0.2 J/K
You need to find the ΔG° for the reaction:
2V(s) + 5CO2(g) -> V2O5(s) + 5CO(g)
To solve this, you can break down the reaction into a combination of the given reactions. Here are the steps to obtain the desired reaction:
1. Multiply Reaction 1 by 2:
2(V2O3(s) + 3CO(g) -> 2V(s) + 3CO2(g))
2. Multiply Reaction 2 by 5:
5(V2O5(s) + 2CO(g) -> V2O3(s) + 2CO2(g))
3. Subtract Reaction 2 from Reaction 1:
2V2O3(s) + 6CO(g) - 5V2O5(s) - 10CO(g) -> 4V(s) + CO2(g)
Notice that we have 10 CO(g) on the left side, so we can use the coefficient of CO2(g) and rewrite the reaction as:
4V(s) + 5CO2(g) - 5V2O5(s) -> 4V(s) + CO2(g)
This simplifies to:
4V(s) + 4CO2(g) - 5V2O5(s) -> CO2(g)
Now, we have the desired reaction equation in terms of the given reactions.
Calculating ΔG° for the desired reaction:
ΔG° = ΔH° - TΔS°
ΔH° = ΔH°(Reaction 1) - 5 * ΔH°(Reaction 2)
= (369.8 kJ) - 5 * (-234.2 kJ)
= 1197.8 kJ
ΔS° = ΔS°(Reaction 1) - 5 * ΔS°(Reaction 2)
= (8.3 J/K) - 5 * (0.2 J/K)
= 7.3 J/K
Temperature, T = 298 K
Now, we can substitute the values into the formula:
ΔG° = ΔH° - TΔS°
= (1197.8 kJ) - (298 K) * (7.3 J/K)
= 1197.8 kJ - 2174 J
= -976.2 kJ - 2.174 kJ
= -978.374 kJ
Therefore, the standard Gibbs free energy change (ΔG°) for the reaction at 298 K is -978.374 kJ.