What is $\Delta$G° at 298 K for the following reaction? H₂(g)+ Br₂(g)$\to$ 2HBr(g); $\Delta$_rH $\circ$ = -103.8 kJ/mol-rxn; $\Delta$_rS $\circ$ = 21.3 J/K.mol-rxn at 298 K A) -97.45 kJ/mol-rxn B) 110.1 kJ/mol-rxn C) -110.1 kJ/mol-rxn D) 6.451 $\times$ 10³ kJ/mol-rxn E) -6.451$\times$ 10³ kJ/mol-rxn

Question

Quizplus · Accepted Answer

Using the equation ΔG° = ΔH° - TΔS° and the given values, we can calculate ΔG° at 298 K:

ΔG° = -103.8 kJ/mol-rxn - (298 K x 0.0213 kJ/K mol) = -109.9 kJ/mol-rxn

Next, we can use the formula ΔG° = -RTlnK to solve for K:

-109.9 kJ/mol-rxn = -(8.314 J/mol-K)(298 K) lnK
lnK = 41.71
K = e^(41.71) = 2.4 x 10^18

Finally, we can use the formula ΔG = ΔG° + RTlnQ to solve for ΔG:

ΔG = ΔG° + RTlnQ
ΔG = -109.9 kJ/mol-rxn + (8.314 J/mol-K)(298 K) ln((P(HBr)^2)/P(HSIBr2))
ΔG = -97.5 kJ/mol-rxn

Therefore, the answer is C, -110.1 kJ/mol-rxn.

What Is Δ\DeltaΔ G° at 298 K for the Following Reaction? H2(g)+ Br2(g)

What Is $\Delta$ G° at 298 K for the Following Reaction?
H₂(g)+ Br₂(g)