What is the enthalpy of combustion, per mole, of butane?

Use [tex]\Delta H_{\text{combustion}} = \sum(\Delta H_{f,\text{ products}}) - \sum(\Delta H_{f,\text{ reactants}})[/tex].

A. [tex]-5315 \text{ kJ/mol}[/tex]
B. [tex]-2657.5 \text{ kJ/mol}[/tex]
C. [tex]2657.4 \text{ kJ/mol}[/tex]
D. [tex]5314.8 \text{ kJ/mol}[/tex]



Answer :

To determine the enthalpy of combustion, per mole, of butane, we use the formula for the enthalpy change of a reaction:

[tex]\[ \Delta H_{reaction} = \sum \left(\Delta H_{f, \text{products}}\right) - \sum \left(\Delta H_{f, \text{reactants}}\right) \][/tex]

Given the options provided:
- [tex]\(-5315 \, \text{kJ/mol}\)[/tex]
- [tex]\(-2657.5 \, \text{kJ/mol}\)[/tex]
- [tex]\(2657.4 \, \text{kJ/mol}\)[/tex]
- [tex]\(5314.8 \, \text{kJ/mol}\)[/tex]

and known typical thermodynamic values, the enthalpy of combustion of butane is usually around [tex]\(-2877 \, \text{kJ/mol}\)[/tex] for complete combustion. Comparing this with the given options, we need to choose the value that is closest to this known typical value.

Among the options given, the value that is closest to [tex]\(-2877 \, \text{kJ/mol}\)[/tex] is:

[tex]\[ -2657.5 \, \text{kJ/mol} \][/tex]

Therefore, the enthalpy of combustion, per mole, of butane is:

[tex]\[ \boxed{-2657.5 \, \text{kJ/mol}} \][/tex]