Pentane (C5H12) is a component of gasoline that burns according to the following

Question

Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12 (l) + 8O2(g) 5CO2 (g) + 6H2O(g) Calculate H°rxn for this reaction using standard enthalpies of formation (Table 10.4). The standard enthalpy of formation of liquid pentane is -146.8 kJ/mol. Need some help thanks show step by step please Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12 (l) + 8O2(g) 5CO2 (g) + 6H2O(g) Calculate H°rxn for this reaction using standard enthalpies of formation (Table 10.4). The standard enthalpy of formation of liquid pentane is -146.8 kJ/mol. Need some help thanks show step by step please Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12 (l) + 8O2(g) 5CO2 (g) + 6H2O(g) Calculate H°rxn for this reaction using standard enthalpies of formation (Table 10.4). The standard enthalpy of formation of liquid pentane is -146.8 kJ/mol. Need some help thanks show step by step please

Explanation / Answer

Delta Hrxn = sum of standard enthalpies of formation of products - sum of enthalpies of formation of reactants.

5 deltaH0f(CO2) + 6 deltaHf0(H2O) - deltaH0f(C5H12) - 8 deltaH0f(O2)

= 5 (-393.5) + 6 (-242) - ( - 146.8) - 8(0)

= - 3272.7 kJ

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