Question

Chemistry question on enthalpy HELP!!!!

Answer 1

Answer: The enthalpy change for this reaction is, -803 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{CH_4}\times \Delta H_{CH_4})]$

where,

n = number of moles  

Now put all the given values in this expression, we get

$\Delta H=[(1\times -394)+(2\times -242]-[(2\times 0)+(1\times -75)]$

$\Delta H=-803kJ$

Therefore, the enthalpy change for combustion of methane is, -803 kJ