At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 6.30 atm, PB = 7.20 atm, PC = 6.40 atm, and PD = 9.10 atm.
2A(g) + 2B(g)-->C(g) + 3D(g)
What is the standard change in Gibbs free energy of this reaction at 25 °C.
Go = -RT*lnK
lnK = p^2*A*p^2*B/(pC* p^3*D
Go = -RT*lnK
K = p^2*A*p^2*B/(pC* p^3*D
K=(pC^1*pD^3)/(pA^2*pB^2)
then go deltaG=-RT*ln(K)
To find the standard change in Gibbs free energy (ΔG°) of a reaction, you can use the equation:
ΔG° = ΔG°f(C) + 3ΔG°f(D) - 2ΔG°f(A) - 2ΔG°f(B)
where ΔG°f(X) represents the standard molar Gibbs free energy of formation for each species X. The values of these standard molar Gibbs free energy of formation can be obtained from a thermodynamic data table or reference book.
Since the reaction is at equilibrium, the standard change in Gibbs free energy (ΔG°) at 25 °C can be calculated using the given equilibrium partial pressures (PA, PB, PC, and PD).
To find the value of ΔG°f(C), ΔG°f(D), ΔG°f(A), and ΔG°f(B), you can use the expressions:
ΔG°f(C) = ΔG°f°(C) + RT ln(PC)
ΔG°f(D) = ΔG°f°(D) + RT ln(PD)
ΔG°°f°(A) + RT ln(PA)
ΔG°f(B) = ΔG°f°(B) + RT ln(PB)
where ΔG°f°(X) represents the standard molar Gibbs free energy of formation at standard conditions (usually 25 °C and 1 atm) for each species X, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin (25 °C + 273.15 = 298.15 K).
Substituting the given values for PA, PB, PC, and PD, and the known values of ΔG°f° for each species, you can calculate the standard change in Gibbs free energy (ΔG°) at 25 °C using the above equation.
Please provide the values of ΔG°f° for C, D, A, and B, so that I can help you calculate the standard change in Gibbs free energy (ΔG°) of this reaction at 25 °C.