2. Calculate the heats of combustion for the following reactions from the standa

Question

2. Calculate the heats of combustion for the following reactions from the standard enthalpies of formation
listed in Appendix 2:
deltaH (C3H8)(g)=-103.9kj/mol
deltaH (H2)(g)= 0 kj/mol
deltaH (C2H6)(g)= -84.7 kj/mol
deltaH(CH4)(g)= -74.85 kj/mol
deltaH(H2S)(g)=-20.15 kj/mol
deltaH(O2)(g)= 0    kj/mol
deltaH(SO2)(g)= -296.1 kj/mol
deltaH(H2O)(l)=   -285.8 kj/mol
deltaH(CL2)(g)= 0   kj/mol
deltaH(HCl)(g)=-92.3   kj/mol

a. C3H8 (g) + H2 (g) --> C2H6 (g) + CH4 (g)
b. 2 H2S (g) + 3 O2 (g) -->2 SO2 (g) + 2 H2O (l)
c. 2 Cl2 (g) + 2 H2O (l) -->2 HCl (g) + O2 (g)

Explanation / Answer

a. C3H8 (g) + H2 (g) --> C2H6 (g) + CH4 (g)

DHrxn = (DHC2H6 + DHCH4) - (DHH2 + DHC3H8)

DHC2H6 = -84.7 kj/mol

DHCH4 = -74.85 kj/mol

DH,H2 = 0

DHC3H8 = -103.9 kj/mol

DHrxn = (-84.7-74.85)-(-103.9) = -55.65 kj

b. 2 H2S (g) + 3 O2 (g) -->2 SO2 (g) + 2 H2O (l)

DHrxn = (2*DHSO2+2*DHH2O) - (2*DHH2S+3*DHO2)

       = (2*-296.1+2*-285.8) - (2*-20.15+3*0)

       = -1123.5 Kj

c. 2 Cl2 (g) + 2 H2O (l) -->2 HCl (g) + O2 (g)

DHrxn = (2*DH,HCl+ DHO2) - (2*DHcl2 + 2*DH,H2O)

       = (2*-92.3 + 0) - (2*0 + 2*-285.8)

       = 387 kj

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