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:

A. –2,657.5 kJ/mol

Explanation:

This is correct on ed-genuity, hope this helps! :)

Answer 2

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

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 :

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

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

$=(8 mol\times -393.5 kJ/mol+10 mol\times (-241.82 kJ/mol))-(2 mol\times (-125.6 kJ/mol)+13 mol\times 0 kJ/mol)=-5315 kJ$

Enthalpy of Combustion of 1 moles of butane :

$\Delta H_c=\frac{5315 kJ}{2mol}=-2657.5 kJ/mol$

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