Question

Consider the following reaction: 2NOBr(g) ? 2NO(g) + Br2(g) K = 0.42 at 373 K. Given that S? of NOBr(g) = 272.6 J/mol·K and that ?S???? and ?H???? are constant with temperature, find the following values. Part 1: ?S???? at 298 K 121.5 J/K Part 2 out of 6 ?G???? at 373 K ? × 10? J/mol Enter your answer in scientific notation.

          Consider the following reaction:
2NOBr(g) ? 2NO(g) + Br2(g)
K = 0.42 at 373 K. Given that S? of NOBr(g) = 272.6 J/mol·K and that ?S???? and ?H???? are constant with temperature, find the following values.

Part 1:
?S???? at 298 K
121.5 J/K

Part 2 out of 6
?G???? at 373 K
? × 10? J/mol
Enter your answer in scientific notation.

Consider the following reaction:
2NOBr(g) ? 2NO(g) + Br2(g)
K = 0.42 at 373 K. Given that S? of NOBr(g) = 272.6 J/mol·K and that ?S???? and ?H???? are constant with temperature, find the following values.

Part 1:
?S???? at 298 K
121.5 J/K

Part 2 out of 6
?G???? at 373 K
? × 10? J/mol
Enter your answer in scientific notation.

Added by Stephanie B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

12/14/2022

Step 1

The ΔS for the reaction is the sum of the ΔS for the products minus the sum of the ΔS for the reactants. Given that ΔS of NOBr(g) = 272.6 J/mol-K, and we have 2 moles of NOBr(g) in the reaction, the total ΔS for the reactants is 2*272.6 = 545.2 J/mol-K. The Show more…

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Consider the following reaction: 2NOBr(g) ⇌ 2NO(g) + Br2(g) K = 0.42 at 373 K. Given that S⁰ of NOBr(g) = 272.6 J/mol·K and that ΔS⁰ᵣ↓ₙ and ΔH⁰ᵣ↓ₙ are constant with temperature, find the following values. Part 1: ΔS⁰ᵣ↓ₙ at 298 K 121.5 J/K Part 2 out of 6 ΔG⁰ᵣ↓ₙ at 373 K J/mol Enter your answer in scientific notation.