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baherus [9]
3 years ago
15

Consider the reaction 2SO2(g) + O2(g) mc021-1.jpg 2SO3(g). Substance mc021-2.jpgHf (kJ/mol) S (J/(molmc021-3.jpgK)) SO3(g) –396

130.58 SO2(g) –297 191.50 O2(g) 0 205.00 What is the mc021-4.jpgGrxn of this reaction, and would it be spontaneous or nonspontaneous at 300.0 K? Use mc021-5.jpgG = mc021-6.jpgH – Tmc021-7.jpgS. –296 kJ/mol, spontaneous –296 kJ/mol, nonspontaneous –100 kJ/mol, spontaneous –100 kJ/mol, nonspontaneous
Chemistry
2 answers:
azamat3 years ago
8 0
This is most likely "-100 kj/mol, spontaneous," but the question is difficult to read
Minchanka [31]3 years ago
6 0

Ans: Spontaneous, -100 kJ/mol

Given reaction:

2SO2(g) + O2(g) → 2SO3(g)

ΔG = ΔH - TΔS -----(1)

ΔH = ∑n(products)ΔH°f(products) - ∑ n(reactants)ΔH°f(reactants)

      = [2ΔH°f(SO3)] - [2ΔH°f(SO2) + ΔH°f(O2)]

      =[2(-396)] - [2(-297) + (0)] = -198 kJ/mol

ΔS = ∑n(products)S°f(products) - ∑ n(reactants)S°f(reactants)

      = [2S°f(SO3)] - [2S°f(SO2) + S°f(O2)]

      =[2(0.13058)] - [2(0.19150) + (0.205)] = -0.32684 kJ/mol.K

Based on equation (1):

ΔG = -198 - (300)(-0.32684) = -99.95 kJ/mol

Ans: ΔG = -100 kJ/mol. Since it is negative, the reaction is spontaneous.

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