For a particular reaction, the equilibrium constant is 1.80 * 10^-² at 316°C, and $\Delta$H° is +12.2 kJ at 25°C. Assuming $\Delta$H° and $\Delta$S° are temperature independent, calculate $\Delta$S° for the reaction.

Question

Quizplus · Accepted Answer

From the equation ΔG°= -RTln(K), we can calculate the standard Gibbs energy change at 316°C as:

ΔG° = ΔH° - TΔS°
ΔG° = 12.2 kJ/mol - (316+273) K * ΔS°
ΔG° = -5.83 kJ/mol

Using this value, we can calculate the equilibrium constant at 316°C as:

ΔG° = -RTln(K)
-ln(K) = ΔG°/RT
-ln(K) = (-5.83 kJ/mol)/(8.314 J/mol*K * 589 K)
ln(K) = 3.33
K = e^(3.33) = 28.1

Now we can use the equilibrium constant to calculate ΔS°:

K = e^(-ΔG°/RT)
28.1 = e^(-ΔG°/(8.314 J/mol*K * 589 K))
ΔG° = -RTln(28.1)
ΔG° = -24.8 kJ/mol
ΔH° = 12.2 kJ/mol

ΔG° = ΔH° - TΔS°
-24.8 kJ/mol = 12.2 kJ/mol - (589 K * ΔS°)
ΔS° = -12.7 J/mol*K

Therefore, the answer is D) -12.7 J/K.

For a Particular Reaction, the Equilibrium Constant Is 1 Δ\DeltaΔ H° Is +122 KJ at 25°C

For a Particular Reaction, the Equilibrium Constant Is 1 $\Delta$ H° Is +122 KJ at 25°C