Question

Butane (c4 h10(g), hf = –125.6 kj/mol) reacts with oxygen to produce carbon dioxide (co2 , hf = –393.5 kj/mol ) and water (h2 o, hf = –241.82 kj/mol) according to the equation below. what is the enthalpy of combustion (per mole) of c4h10 (g)? use . –2,657.5 kj/mol –5315.0 kj/mol –509.7 kj/mol –254.8 kj/mol

Answer 1

Answer: -2657.5 kJ/mol. (A)

Explanation: took the test ;D

Answer 2

Answer: The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.

Step-by-step explanation:

`\Delta H_(f,CO_2)=-393.5 kJ/mol`

`\Delta H_(f,H_2O)=-241.82 kJ/mol`

`\Delta H_{f,C_4H_(10)}=-125.6 kJ/mol`

`\Delta H_(f,O_2)=0 kJ/mol`

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

Enthalpy of Combustion of 2 moles of butane :

`=\sum(\Delta H_f\text{of products})-\sum(\Delta H_f\text{of reactants})`

`\Delta H_c=(8\Delta H_(f,CO_2)+10\Delta H_(f,H_2O))-(2\Delta H_{f,C_4H_(10)}-13\Delta H_(f,O_2))`

`=(8 mol* -393.5 kJ/mol+10 mol* (-241.82 kJ/mol))-(2 mol* (-125.6 kJ/mol)+13 mol* 0 kJ/mol)=-5315 kJ`

Enthalpy of Combustion of 1 moles of butane :

`\Delta H_c=(5315 kJ)/(2mol)=-2657.5 kJ/mol`

The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.