Answer: The enthalpy change for this reaction is, -803 kJ
Explanation:
The balanced chemical reaction is,
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{CH_4}\times \Delta H_{CH_4})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(1\times -394)+(2\times -242]-[(2\times 0)+(1\times -75)][/tex]
[tex]\Delta H=-803kJ[/tex]
Therefore, the enthalpy change for combustion of methane is, -803 kJ
