Question HLPaper 1AAnswerWhat is the enthalpy change of the reaction? C6H14(l)→C2H4(g)+C4H10(g)C_6H_{14} (l) → C_2H_4 (g) + C_4H_{10} (g)C6H14(l)→C2H4(g)+C4H10(g) ΔHcombustion⊖(C6H14(l))=−4163 kJ mol−1\Delta H^\ominus_\text{combustion} (C_6H_{14} (l))=-4163 \ kJ \ mol^{-1}ΔHcombustion⊖(C6H14(l))=−4163 kJ mol−1 ΔHcombustion⊖(C2H4(g))=−1411 kJ mol−1\Delta H^\ominus_\text{combustion} (C_2H_4 (g))=-1411 \ kJ \ mol^{-1}ΔHcombustion⊖(C2H4(g))=−1411 kJ mol−1 ΔHcombustion⊖(C4H10(g))=−2878 kJ mol−1\Delta H^\ominus_\text{combustion} (C_4H_{10} (g))=-2878 \ kJ \ mol^{-1}ΔHcombustion⊖(C4H10(g))=−2878 kJ mol−1A+ 1411 − 2878 − 4163B+ 1411 + 2878 − 4163C− 1411 − 2878 + 4163D+ 1411 + 2878 + 4163
