1 point Consider the following equation: S(s, rhombic ) + O2( g) → SO2( g) At 298 K, AH° for this reaction is -269.9 kJ/mol, and AS° is +11.6 J/K. What is the value of AG° at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide? -300.4 +300.4 -4,597 +4,597 -274.2
1 point Consider the following equation: S(s, rhombic ) + O2( g) → SO2( g) At 298 K, AH° for this reaction is -269.9 kJ/mol, and AS° is +11.6 J/K. What is the value of AG° at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide? -300.4 +300.4 -4,597 +4,597 -274.2
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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From attempting to find ΔG I’ve used the equation of ΔG=ΔH-TΔS. I found the change in temp and changed the units of ΔS to kJ. The answer that I keep getting is -270.77. Which is not an answer listed and I’m not sure where I’m going wrong. Please help me find my error
![1 point
Consider the following equation:
S(s, rhombic ) + O2( g) → SO2( g)
At 298 K, AH° for this reaction is -269.9 kJ/mol, and AS° is +11.6 J/K.
What is the value of AG° at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide?
-300.4
+300.4
-4,597
+4,597
-274.2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8be9766c-e34f-4cb9-8166-5f982ecec658%2Fa4ad1372-181c-476d-b251-afe268b36d85%2Fwdgaiuf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1 point
Consider the following equation:
S(s, rhombic ) + O2( g) → SO2( g)
At 298 K, AH° for this reaction is -269.9 kJ/mol, and AS° is +11.6 J/K.
What is the value of AG° at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide?
-300.4
+300.4
-4,597
+4,597
-274.2
Expert Solution
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Step 1
We know the thermodynamic equation.,
∆G° = ∆H° - T∆S°
where,
∆H° = -269.9 KJ/mol
∆S° = +11.6 J/mol.K = +0.0116 KJ/mol.K
(1 J = 0.001 KJ then 11.6 J = 11.6 X 0.001 = 0.0116 KJ)
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