Question

Butane C4 H10 (g),(Delta.Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2 O(g) (Delta.Hf = –241.82) in the reaction:

2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper O subscript 2 (g) right arrow 8 upper C upper O subscript 2 plus 10 upper H subscript 2 upper O (g).

What is the enthalpy of combustion, per mole, of butane?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–5,314.8 kJ/mol
–2,657.4 kJ/mol
2,657.4 kJ/mol
5,314.8 kJ/mol

Answer 1

B: –2,657.4 kJ/mol

Step-by-step explanation:

took the test

Answer 2

• Second choice:

`\Delta H_(rxn)=-2,657.4kJ/mol`

Step-by-step explanation:

1. Combustion reaction

`2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O`

2. Enthalpy of reaction equation

`\Delta H_(rxn)=\sum\Delta H_f(products)-\sum\Delta H_f(reactants)`

3. Substitute and divide by the coefficient of butane that appears in the combustion reaction

`\Delta H_(rxn)=(8(-393.5kJ/mol)+10(-241.82kJ/mol)-2(-125.7kJ/mol))/(2mol)`

`\Delta H_(rxn)=-2,657.4kJ/mol\leftarrow answer`