For the reaction: CO(g)+ 2 H₂(g)→ CH₃OH(g) K_p = 91.4 at 350 K and K_p = 2.05 × 10^-4 at 298 K. What is the value of Δ_rH°? A)49.9 kJ/mol B)2.08 × 10³ kJ/mol C)3.74 × 10^-2 kJ/mol D)217 kJ/mol E)446 kJ/mol

Question

Quizplus · Accepted Answer

From the given equilibrium constant values, we can calculate ΔG° at 298 K as follows:
ΔG° = -RTlnKp
ΔG° = -8.314 J/mol·K × (ln2.05 × 10^-4) = 217 kJ/mol
To calculate ΔS°, we need to use the equation:
ΔG° = ΔH° - TΔS°
ΔS° = (ΔH° - ΔG°)/T
Since we do not have the value of ΔH°, we cannot calculate ΔS° directly. However, we know that ΔS° is positive if the reaction produces more gas molecules than it consumes, and negative if the reaction consumes more gas molecules than it produces. In this case, we can see that the reaction consumes one molecule of CO and two molecules of H2S, but produces one molecule of CH3OH. Therefore, we expect ΔS° to be negative. Among the answer choices, only D is negative, so D is the correct answer.

For the Reaction: CO(g)+ 2 H2(g)→ CH3OH(g)

For the Reaction: CO(g)+ 2 H₂(g)→ CH₃OH(g)