Sapling Learning University of My Assignment Attempts Score 50 33 Copyright 20 O

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Sapling Learning University of My Assignment Attempts Score 50 33 Copyright 20 O Logout Help Jump to Activities and Due Dates Chap CHM 2046 5 ALL SECTIONS Resources 2/2015 11:30 PM A 71.2/100 Gradebook O Assignment Information Print A calculator Periodic Table Available From: Not S Question 11 of 11 CO 11/12/2015 11:30 PM Due Date: Map Points Possible: 000 Grade Category: Graded Substance (kJ/mol) Consider the decomposition of a metal oxide to its CHM 2046 Fall 15 elements, where M represents a generic metal. Policies: M203 3610 n check your answers. n view solutions when you complete or give up on any question What is the standard change in Gibbs energy for the reaction, as written n the forward direction? You have ten attempts per question Number There is no penalty for incorrect answers. AG 6.10 kJ mo o Help With This Top What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? Web Help & Video O Technical Support and Bug Reports What is the equilibrium pressure of O2(g) over M(s) at 298 K? ber 64 atm Next Previous Give Up & View Solution Check Answer Hint

Explanation / Answer

Given that is

M2O3 (s) <------> 2M(s) + 3/2 O2 (g)

Gorxn

Gfo [M2O3(s)] = -6.10 kJ/mol

Gfo [M(s)] = 0 kJ/mol

Gfo (O2) = 0 kJ/mol

Gorxn = Gfo(products) - Gfo( reactants)

= 2Gfo [M(s)] + (3/2)Gfo (O2) - 3/2Gfo [M2O3(s)]  

= 2 (0) kJ/mol + (3/2) [0 kJ/mol] - (3/2) [ -6.10 kJ/mol]

= + 9.15 kJ/mol

Therefore, Gorxn = + 9.15 kJ/mol

K

Gorxn = -nRT ln K where R = 8.314 J/mol/K

Gorxn = + 9.15 kJ/mol = 9150 J/mol

T = 298 K

K = e -(Gorxn/RT)  

= e -(9150 / 8.314 x x 298)

= 0.025

Therefore, K = 0.025

PO2

Given reaction M2O3 (s) <------> 2M(s) + 3/2 O2 (g)

K = (PO2) 3/2 (solids cannot be taken into consideration)

0.025 = (PO2) 3/2   

PO2  =   (0.025)2/3

=0.085 atm

Hence, PO2 = 0.085 atm

Hence, equilibrium pressure of O2(g) over M(s) at 298 K = 0.085 atm

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