Question

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Answer 1

Answer: The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ