Question

Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, graphite and hydrogen are -890 kJ mol",-393.4 kimol, and -285.7 kJ moll.

Answer 1

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of
`CH_4` will be,

`C(s)+2H_2(g)\rightarrow CH_4(g)`
`\Delta H_(formation)=?`

The intermediate balanced chemical reaction will be,

(1)
`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)`
`\Delta H_1=-890kJ/mole`

(2)
`C(s)+O_2(g)\rightarrow CO_2(g)`
`\Delta H_2=-393.4kJ/mole`

(3)
`H_2(g)+(1)/(2)O_2(g)\rightarrow H_2O(l)`
`\Delta H_3=-285.7kJ/mole`

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1)
`CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)`
`\Delta H_1=890kJ/mole`

(2)
`C(s)+O_2(g)\rightarrow CO_2(g)`
`\Delta H_2=-393.4kJ/mole`

(3)
`2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)`
`\Delta H_3=2* (-285.7kJ)=-571.4kJ/mole`

The expression for enthalpy of formation of
`CH_4` will be,

`\Delta H_(formation)=\Delta H_1+\Delta H_2+\Delta H_3`

`\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)`

`\Delta H=-74.8kJ/mole`

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole