Answer:
To calculate Δ∘rxn, we can use the following formula:
ΔG∘rxn = ΔH∘rxn - TΔS∘rxn
where ΔH∘rxn is the enthalpy change of the reaction, T is the temperature in Kelvin, and ΔS∘rxn is the entropy change of the reaction.
We know that ΔH∘rxn = -44.2 kJ and we want to find ΔS∘rxn at 25.0 ∘C (298 K). We can use the following formula to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK
where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.
We can find K using the following formula:
ΔG∘rxn = -RTlnK K = e^(-ΔG∘rxn/RT)
We know that ΔG∘rxn = -44.2 kJ/mol and R = 8.314 J/mol K, so we can calculate K:
K = e^(-(-44.2 kJ/mol)/(8.314 J/mol K * 298 K)) K = 1.9 x 10^7
Now we can use K to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK ΔS∘rxn = -(ΔH∘rxn - ΔG∘rxn)/T ΔS∘rxn = -((-44.2 kJ/mol) - (-8.314 J/mol K * 298 K * ln(1.9 x 10^7)))/(298 K) ΔS∘rxn = -0.143 kJ/K
Therefore, ΔS∘rxn is -0.143 kJ/K.
To determine whether the reaction is spontaneous at 25 ∘C and standard pressure, we can use Gibbs free energy (ΔG). If ΔG < 0, then the reaction is spontaneous in the forward direction; if ΔG > 0, then it is spontaneous in the reverse direction; if ΔG = 0, then it is at equilibrium.
We know that ΔG∘rxn = -44.2 kJ/mol and T = 25 ∘C (298 K). We can use the following formula to calculate ΔG:
ΔG = ΔG∘ + RTlnQ
where Q is the reaction quotient.
At equilibrium, Q = K (the equilibrium constant). Since we calculated K earlier to be 1.9 x 10^7, we can use this value for Q.
ΔG = ΔG∘ + RTlnQ ΔG = (-44.2 kJ/mol) + (8.314 J/mol K * 298 K * ln(1.9 x 10^7)) ΔG = -43.6 kJ/mol
Since ΔG < 0, the reaction is spontaneous in the forward direction at 25 ∘C and standard pressure.