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

H°rxn=mHf°(products)nHf°(reactants)∆H°_{rxn} = \sum m ∆H^°_f (products) - \sum n ∆H^°_f (reactants)

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

Hf°∆H^°_f is the standard heat of formation.

Hrxn°=[2  mol×(296.8  kJ/mol)+2  mol×(241.8  kJ/mol)][2  mol×(20.2  kJ/mol)+3  mol×0]=[593.6  kJ483.6  kJ][40.4  kJ]=1036.8  kJ∆H^°_{rxn} = [2 \;mol \times (-296.8 \;kJ/mol) + 2 \;mol \times (-241.8 \;kJ/mol)] – [2 \;mol \times (-20.2 \;kJ/mol) + 3\; mol \times 0] \\ = [-593.6 \;kJ -483.6 \;kJ] -[-40.4 \;kJ] \\ = -1036.8 \;kJ

(b) Balanced equation:

CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)

Hrxn°=[1  mol×(139  kJ/mol)+4×(92.3  kJ/mol)][1  mol×(74.9  kJ/mol)+4  mol×0]=[139  kJ369.2  kJ][74.9  kJ]=433  kJ∆H^°_{rxn} = [1 \;mol \times (-139 \;kJ/mol) + 4 \times (-92.3 \;kJ/mol)] – [1 \;mol \times (-74.9\;kJ/mol) + 4 \;mol \times 0] \\ = [-139\; kJ -369.2 \;kJ] -[-74.9 \;kJ] \\ = -433 \;kJ


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