Answer to Question #242192 in General Chemistry for KING

Question #242192

. Calculate ∆H°rxn for each of the following:

(a) 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O (g)

(b) CH4(g) + CI2(g) → CCI4(l) + HCl(g) [unbalanced


1
Expert's answer
2021-09-26T12:14:20-0400

(a) 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O (g)

Use the following equation to determine the standard heat of reaction

"\u2206H\u00b0_{rxn} = \\sum m \u2206H^\u00b0_f (products) - \\sum n \u2206H^\u00b0_f (reactants)"

m and n are the number of moles of products and reactants, respectively.

"\u2206H^\u00b0_f" is the standard heat of formation.

"\u2206H^\u00b0_{rxn} = [2 \\;mol \\times (-296.8 \\;kJ\/mol) + 2 \\;mol \\times (-241.8 \\;kJ\/mol)] \u2013 [2 \\;mol \\times (-20.2 \\;kJ\/mol) + 3\\; mol \\times 0] \\\\\n\n= [-593.6 \\;kJ -483.6 \\;kJ] -[-40.4 \\;kJ] \\\\\n\n= -1036.8 \\;kJ"

(b) Balanced equation:

CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)

"\u2206H^\u00b0_{rxn} = [1 \\;mol \\times (-139 \\;kJ\/mol) + 4 \\times (-92.3 \\;kJ\/mol)] \u2013 [1 \\;mol \\times (-74.9\\;kJ\/mol) + 4 \\;mol \\times 0] \\\\\n\n= [-139\\; kJ -369.2 \\;kJ] -[-74.9 \\;kJ] \\\\\n\n= -433 \\;kJ"


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