At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 4.10 bar, PB = 5.60 bar, PC = 1.20 bar, and PD = 9.70 bar.
2A(g)+2B(g)-->C(g)+ 3D(g)
What is the standard change in Gibbs free energy of this reaction at 25 °C.
I don't really get how to even start the problem because I never seen partial pressure in the Gibb free energy. I just know that delta G = -RTlnK.
You have the equilibrium pressures, calculate Kp, then
dGo = -R*T*lnK
To determine the standard change in Gibbs free energy (ΔG°) for this reaction at 25 °C, we need to find the equilibrium constant (K) first. The equilibrium constant can be obtained from the given equilibrium partial pressures using the ideal gas law.
The ideal gas law states that PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we have partial pressures (P) for the reactants and products, as well as the given temperature (25 °C = 25 + 273 = 298 K). The ideal gas law can be rearranged to solve for n/V (moles per unit volume):
n/V = P/RT
Now, let's calculate the moles per unit volume for each substance:
For substance A:
nA/V = PA/RT = 4.10 bar / (0.0831 bar⋅L⋅mol^(-1)⋅K^(-1) ⋅298 K) = 0.1669 mol/L
For substance B:
nB/V = PB/RT = 5.60 bar / (0.0831 bar⋅L⋅mol^(-1)⋅K^(-1) ⋅298 K) = 0.2292 mol/L
For substance C:
nC/V = PC/RT = 1.20 bar / (0.0831 bar⋅L⋅mol^(-1)⋅K^(-1) ⋅298 K) = 0.049 mol/L
For substance D:
nD/V = PD/RT = 9.70 bar / (0.0831 bar⋅L⋅mol^(-1)⋅K^(-1) ⋅298 K) = 0.3969 mol/L
Now, let's use the stoichiometric coefficients of the balanced equation to calculate the reaction quotient (Q) using the expression:
Q = [C]^c[D]^d / ([A]^a[B]^b)
For this reaction:
Q = [C]^1[D]^3 / ([A]^2[B]^2) = (nC/V)^1 * (nD/V)^3 / ((nA/V)^2 * (nB/V)^2)
Substituting the calculated values:
Q = (0.049 mol/L)^1 * (0.3969 mol/L)^3 / ((0.1669 mol/L)^2 * (0.2292 mol/L)^2) = 3.517
Now, with Q, we can calculate the ΔG° using the equation:
ΔG° = -RT ln(K)
Since the reaction is at equilibrium, Q = K, so
ΔG° = -RT ln(Q) = -(8.314 J/(mol⋅K)) * 298 K * ln(3.517) = -5,915 J/mol
Therefore, the standard change in Gibbs free energy (ΔG°) for this reaction at 25 °C is -5,915 J/mol.