Sho Feedback6 of 15Review I Constants I Periodic TablePart AFind AGrxn when PNO,=0.103 atm and PNO = 8.0 atm at 25 °C.3 NO2(g) + H20(1) → 2 HNO3(1) + NO(g)-12.6 kJ2.20 x 104 kJO 19.0 kJ31.4 kJ

Since we know that ΔGrxn = ΔG0rxn + RTln(K) where R = gas constant = 8.314 and T = temperature in…

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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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Question

Sho
<CHE154 S20 Ch19 Sec8-10
Self Assessment 19.10 Enhanced-with Feedback
MISSED THIS? Read Section 19.9
O.
(Pages 878-880) ; Watch IWE 19.10 .
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The following reaction, AGxn =
9.4kJ at 25 °C.
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IS

Feedback
6 of 15
Review I Constants I Periodic Table
Part A
Find AGrxn when PNO,
=0.103 atm and PNO = 8.0 atm at 25 °C.
3 NO2(g) + H20(1) → 2 HNO3(1) + NO(g)
-12.6 kJ
2.20 x 104 kJ
O 19.0 kJ
31.4 kJ

Expert Solution

Step 1

Since we know that

ΔG_rxn = ΔG⁰_rxn + RTln(K)

where R = gas constant = 8.314

and T = temperature in Kelvin = 25 + 273 = 298 K

K = equilibrium constant

ΔG⁰_rxn in J = 9400 J

For the given reaction,

K = P_NO / (P_NO2 )³

Hence substituting the value we get

K = 8 / 0.103³ = 7321.13

Step by step

Solved in 2 steps

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