Consider the decomposition of a metal oxide to its elements, where Mrepresents a generic metal.SubstanceAG°{(kJ/mol)1M,0(s) 2 M(s) +0,(g)2M20(s)-7.60M(s)What is the standard change in Gibbs energy for the reaction, as written, in theO2(g)forward direction?J/molAGixn =contact ushelpterms of useabout usprivacy po icy6:12 PMcareers3/25/2021insert144&8.行 What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?K =What is the equilibrium pressure of O2(g) over M(s) at 298 K?atmPo,contact ushelp6:13 Ftems of useprivacy po lcyabout uscareers3/25/2

Equillibrium pressure is considered only for gas in heterogeneous Equillibrium.

eric metal. AG°(kJ/mol) Substance 1 M20(s) -7.60 E 2 M(s) + 0,(g) M(s) 01 O2(g) 0. ndard change in Gibbs energy for the reaction, as written, in the on? kl/mol contact us help 6:12 PM terms of use privacy polcy 3/25/2021 careers

eric metal. AG°(kJ/mol) Substance 1 M20(s) -7.60 E 2 M(s) + 0,(g) M(s) 01 O2(g) 0. ndard change in Gibbs energy for the reaction, as written, in the on? kl/mol contact us help 6:12 PM terms of use privacy polcy 3/25/2021 careers

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

Consider the decomposition of a metal oxide to its elements, where M
represents a generic metal.
Substance
AG°{(kJ/mol)
1
M,0(s) 2 M(s) +0,(g)
2
M20(s)
-7.60
M(s)
What is the standard change in Gibbs energy for the reaction, as written, in the
O2(g)
forward direction?
J/mol
AGixn =
contact us
help
terms of use
about us
privacy po icy
6:12 PM
careers
3/25/2021
insert
144
&
8.
行

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?
K =
What is the equilibrium pressure of O2(g) over M(s) at 298 K?
atm
Po,
contact us
help
6:13 F
tems of use
privacy po lcy
about us
careers
3/25/2

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2) Calculate the numeric value of the equilibrium constant K for reaction at 298 K: 2 NO2 (g) → N₂O4 (g) from Data Section: les of das Athina.pdf-Adebe Acrobat Reader (54-b0) You SignWindow H Tools Home 827-31.40 - Tables of data Atkin x a Table 2.7 (Continued) Neon Ne(g) Nitrogen N₂(g) N(g) NO(g) N₂O(g) NO₂(g) N₂O₂(g) N₂O5(s) N₂O,(g) 1081 (12 of 4) TTATO (11 A M/(g mol-¹) 20.18 28.013 14.007 30.01 44.01 46.01 92.1 108.01 108.01 OO 201 8. P AH*/(kJ mol-¹) 0 0 +472.70 +90.25 +82.05 +33.18 +9.16 -43.1 +11.3 174 17 AG/(kJ mol-¹) 0 0 +455.56 +86.55 +104.20 +51.31 +97.89 +113.9 +115.1 ℗ 30 31 x Sign In &00 S B D 2 C %
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For a certain chemical reaction, the standard Gibbs free energy of reaction at 25.0 °C is 95.6 kJ. Calculate the equilibrium constant K for this reaction. Round your answer to 2 significant digits. K = || 国 Explanation Check O 2021 McGraw Hill Education. All Rights Reserved Terms of Use Privacy Accenibility APR 30 MacBook Air
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Under suitable conditions the reaction 2NO(g) +N2O4(g) -> 2N2O3(g) can be studied in the gas phase. Fromthe data below, calculate Gibbs free energy change for the reaction at 298 K. NO (g) 90.29 (H) 210.7 (S) N2O4 (g) 9.08 (H) 304.3 (S) N2O3 (g) 82.84 (H) 309.3 (S)