Bb Thread: Discussion 6 20 X Parker University CHEM X Search d/ibis/ pling 8/4/2

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Bb Thread: Discussion 6 20 X Parker University CHEM X Search d/ibis/ pling 8/4/2016 10:55 PM A 18.3/20 8/4/2016 07:08 PM Gradebook s Score Print Calculator Periodic Table 100 Question 13 of Map 100 Sapling Learning 100 For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction 100 2No, (g) 2NOI 100 the standard change in Gibbs free energy is AG 69.0 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are 95 PNo. 0.250 bar, Po, 0.200 bar, and PNO 0.900 bar 97 Number kJ mol AG 98 100 Previous Give Up & View Solution Check Answer Nex Exit Hin 100 The change in Gibbs free energy is given by 95 AGE AG o RTin (Q) 100 where AG is the change in Gibbs free energy, AG is the standard change in Gibbs free energy, R is the ideal-gas constant 8.3145 J/(mol. K), Tis absolute temperature, and Q is the thermodynamic reaction quotient. Assignment In Available From: Due Date: Points Possible: Grade Category: Description: Policies: You can check you You can view solu up on any questic You can keep try you get it right or You lose 5% of the f in your question answer. O eTextbook O Help With This 8:42 PM 8/4/2016

Explanation / Answer

Solution:- reaction quotient, Q = [partial pressure of products]/[partial pressure of reactants]

Q = [partial pressure of NO2]2/([partial pressure of NO]2[partial pressure of O2])

Plug in the values of partial pressures of each gas in the equation.

Q = (0.900)2/(0.250)2(0.200)

Q = 64.8

we know that, delta G = delta G0 + RT ln[Q]

where R is gas constant and its value in terms of kJ is 8.314 x 10-3 kJ mol-1 K-1.

Plug in the values in it..

delta G = -69.0 kJ mol-1 + 8.314 x 10-3 kJ mol-1 K-1 x 298 K x ln(64.8)

delta G = -69.0 kJ mol-1 + 10.3 kJ mol-1

delta G = -58.7 kJ mol-1 or -58.7 kJ/mol

So, the answer is -58.7 kJ/mol.

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