Question

14. Calculate the standard Gibbs free energy change, ?G° (in kJ mol?¹), at a temperature of 425 K, for the reaction, 2 CO(g) + O?(g) ? 2 CO?(g) at 298 K: CO(g) O?(g) CO?(g) ?H?° (kJ mol?¹) -110.5 0 -393.5 S° (J mol?¹K?¹) 198 205 214 ?G?° (kJ mol?¹) -137 0 -394 A) -514 B) -566.0 C) +72.9 D) -639.5 E) -492

          14. Calculate the standard Gibbs free energy change, ?G° (in kJ mol?¹), at a temperature of 425 K, for the reaction,
2 CO(g) + O?(g) ? 2 CO?(g)
at 298 K:
CO(g) O?(g) CO?(g)
?H?° (kJ mol?¹) -110.5 0 -393.5
S° (J mol?¹K?¹) 198 205 214
?G?° (kJ mol?¹) -137 0 -394
A) -514
B) -566.0
C) +72.9
D) -639.5
E) -492

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14. Calculate the standard Gibbs free energy change, ?G° (in kJ mol?¹), at a temperature of 425 K, for the reaction,
2 CO(g) + O?(g) ? 2 CO?(g)
at 298 K:
CO(g) O?(g) CO?(g)
?H?° (kJ mol?¹) -110.5 0 -393.5
S° (J mol?¹K?¹) 198 205 214
?G?° (kJ mol?¹) -137 0 -394
A) -514
B) -566.0
C) +72.9
D) -639.5
E) -492

Added by Lauren S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

09/02/2023

Step 1

ΔH = ΣΔH(products) - ΣΔH(reactants) ΔH = 2 * (-393.5) - (2 * (-110.5)) ΔH = -787 + 221 ΔH = -566 kJ/mol ΔS = ΣΔS(products) - ΣΔS(reactants) ΔS = 2 * 214 - (2 * 198 + 205) ΔS = 428 - 601 ΔS = -173 J/mol*K Now, we can calculate the standard Gibbs free energy Show more…

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Calculate the standard Gibbs free energy change, ΔG° (in kJ mol⁻¹), at a temperature of 425 K, for the reaction, 2 CO(g) + O₂(g) → 2 CO₂(g) at 298 K: CO(g) O₂(g) CO₂(g) ΔHፐ° (kJ mol⁻¹) -110.5 0 -393.5 S° (J mol⁻¹K⁻¹) 198 205 214 ΔGፐ° (kJ mol⁻¹) -137 0 -394

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