Question

Chemistry question on enthalpy HELP!!!!

Answer 1

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

Step-by-step 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* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(CO_2)* \Delta H_(CO_2))+(n_(H_2O)* \Delta H_(H_2O))]-[(n_(O_2)* \Delta H_(O_2))+(n_(CH_4)* \Delta H_(CH_4))]`

where,

n = number of moles

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

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

`\Delta H=-803kJ`

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