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 the given 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:
ΔH° for the first reaction = 369.8 kJ
ΔS° for the first reaction = 8.3 J/K
ΔH° for the second reaction = -234.2 kJ
ΔS° for the second reaction = 0.2 J/K
Temperature (T) = 298 K
Step 1: Convert the units of ΔH° and ΔS° for the second reaction to kJ and kJ/K, respectively.
ΔH° for the second reaction = -234.2 kJ
ΔS° for the second reaction = 0.2 kJ/K
Step 2: Calculate ΔG° for the first reaction.
ΔG°1 = ΔH°1 - TΔS°1
= 369.8 kJ - (298 K)(8.3 kJ/K)
= 369.8 kJ - 2473.4 kJ
= -2103.6 kJ
Step 3: Calculate ΔG° for the second reaction.
ΔG°2 = ΔH°2 - TΔS°2
= -234.2 kJ - (298 K)(0.2 kJ/K)
= -234.2 kJ - 59.6 kJ
= -293.8 kJ
Step 4: Calculate ΔG° for the overall reaction by subtracting the ΔG° of the second reaction from that of the first reaction.
ΔG°overall = ΔG°1 - ΔG°2
= -2103.6 kJ - (-293.8 kJ)
= -2103.6 kJ + 293.8 kJ
= -1809.8 kJ
Therefore, the standard Gibbs free energy change (ΔG°) for the given reaction at 298 K is -1809.8 kJ.
To calculate the standard Gibbs free energy change (ΔG°) for the given reaction, you can use the standard Gibbs free energy change equation:
ΔG° = ΔH° - TΔS°
where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.
For the reaction, 2V(s) + 5CO2(g) -> V2O5(s) + 5CO(g), we need to find the ΔG°.
1. Calculate the ΔH° for the reaction:
The ΔH° of the reaction will be the sum of the ΔH° values for each reactant and product.
ΔH° = ΔH°(V2O5) + ΔH°(5CO) - ΔH°(2V) - ΔH°(5CO2)
ΔH°(V2O5) = ΔH°(formation of V2O5) = ΔH°(V2O5) - ΔH°(V2O3)
= -234.2 kJ - (369.8 kJ/2) = -234.2 kJ - 184.9 kJ = -419.1 kJ
ΔH°(5CO) = 5 * ΔH°(CO) = 5 * ΔH°(CO) - 5 * ΔH°(CO2)
= 5 * 0 kJ/mol - 5 * (-393.5 kJ/mol) = +1967.5 kJ
ΔH°(2V) = 2 * ΔH°(V(s)) = 2 * ΔH°(V(s)) - 2 * ΔH°(V2O3)
= 2 * 0 kJ/mol - 2 * 369.8 kJ/mol = -739.6 kJ
ΔH°(5CO2) = 5 * ΔH°(CO2(g)) = 5 * ΔH°(CO2(g)) - 5 * ΔH°(CO(g))
= 5 * (-393.5 kJ/mol) - 5 * 0 kJ/mol = -1967.5 kJ
ΔH° = -419.1 kJ + 1967.5 kJ + (-739.6 kJ) + (-1967.5 kJ) = -158 kJ
2. Calculate the ΔS° for the reaction:
The ΔS° of the reaction will be the sum of the ΔS° values for each reactant and product.
ΔS° = ΔS°(V2O5) + ΔS°(5CO) - ΔS°(2V) - ΔS°(5CO2)
ΔS°(V2O5) = ΔS°(formation of V2O5) = ΔS°(V2O5) - ΔS°(V2O3)
= 0.2 J/(mol·K) - 8.3 J/(mol·K) = -8.1 J/(mol·K)
ΔS°(5CO) = 5 * ΔS°(CO) = 5 * ΔS°(CO) - 5 * ΔS°(CO2)
= 5 * 0 J/(mol·K) - 5 * 0 J/(mol·K) = 0 J/(mol·K)
ΔS°(2V) = 2 * ΔS°(V(s)) = 2 * ΔS°(V(s)) - 2 * ΔS°(V2O3)
= 2 * 0 J/(mol·K) - 2 * 8.3 J/(mol·K) = -16.6 J/(mol·K)
ΔS°(5CO2) = 5 * ΔS°(CO2(g)) = 5 * ΔS°(CO2(g)) - 5 * ΔS°(CO(g))
= 5 * 0 J/(mol·K) - 5 * 8.3 J/(mol·K) = -41.5 J/(mol·K)
ΔS° = -8.1 J/(mol·K) + 0 J/(mol·K) + (-16.6 J/(mol·K)) + (-41.5 J/(mol·K)) = -66.2 J/(mol·K)
3. Calculate ΔG° using the equation:
Remember to convert the temperature from Celsius to Kelvin by adding 273.15.
ΔG° = ΔH° - TΔS°
= -158 kJ - (298 K * (-66.2 J/(mol·K)) * (1 kJ/1000 J))
= -158 kJ - (-19.7 kJ)
= -138.3 kJ
Therefore, the ΔG° for the reaction 2V(s) + 5CO2(g) -> V2O5(s) + 5CO(g) at 298 K is -138.3 kJ.