At 699 K,$\Delta$G° = -23.25 kJ for the reaction H₂(g)+ I₂(g) 2HI(g).Calculate $\Delta$G for this reaction if the reagents are both supplied at 10.0 atm pressure and the product is at 1.76 atm pressure.

Question

Quizplus · Accepted Answer

We can use the equation:

ΔG = ΔG° + RTln(Q)

where ΔG is the free energy change under non-standard conditions, ΔG° is the free energy change under standard conditions, R is the gas constant, T is the temperature in kelvin, and Q is the reaction quotient.

First, let's calculate Q:

Q = (PH2I)^2 / (PHS2UB12S2UB0)(PIS2UB12S2UB0)

Q = (10.0 atm)^2 / (1.76 atm)^2 ≈ 322.57

Now, let's plug in the values:

ΔG = (-23.25 kJ/mol) + (8.314 J/mol-K)(699 K)(ln(322.57))

ΔG ≈ -43.4 kJ/mol

Therefore, the answer is D. The negative sign indicates that the reaction is exothermic.

At 699 K Δ\DeltaΔ G° = -2325 KJ for the Reaction H2(g)+ I2(g) 2HI(g)

At 699 K $\Delta$ G° = -2325 KJ for the Reaction H₂(g)+ I₂(g) 2HI(g)