Consider the reaction 2SO₂(g) + O₂(g) 2SO₃(g) For which $\Delta$H° = -200. kJ and $\Delta$S° = -187 J/K at 25°C. Assuming that $\Delta$H° and $\Delta$S° are independent of temperature, calculate the temperature where K_p = 1.

Question

Quizplus · Accepted Answer

We can use the equation:

ΔG° = -RTln(K)

Where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

From the given information, we know:

ΔH° = -200 kJ/mol
ΔS° = -187 J/K

Converting ΔH° to J/mol:

ΔH° = -200000 J/mol

Substituting these values into the equation:

ΔG° = (-(200000 J/mol))/(8.314 J/mol*K)*(T) + (-187 J/K)*(ln(K))

At equilibrium, ΔG° = 0, so:

0 = (-(200000 J/mol))/(8.314 J/mol*K)*(T) + (-187 J/K)*(ln(1))

Solving for T:

T = (-(0 J/K) - (-187 J/K)*(ln(1)))/((-(200000 J/mol))/(8.314 J/mol*K))

T = 1070 K

Therefore, the answer is B) 1070 K.

Consider the Reaction 2SO2(g) + O2(g) 2SO3(g) for Which Δ\DeltaΔ

Consider the Reaction 2SO₂(g) + O₂(g) 2SO₃(g)
for Which $\Delta$