Question

Enthalpy of sublimation of graphite is \( 716 kJ mol ^{-1} \). \begin{tabular}{|l|c|c|} \hline Bond enthalpy & \( H - H \) & \( C - H \) \\ \hline\( \Delta H ^{0} kJ \textrm {mol } ^ { - 1 } \) & \( 4 3 6 . 4 \) & \( 4 1 4 \) \\ \hline \end{tabular} Calculate standard enthalpy of formation of \( CH _{4} \). (3 marks)

Answer 1

We have:

Δ_subH⁰_graphite = 716 kJ mol^-1

Bond enthalpy	H - H	C - H
ΔH0 kJ mol-1	436.4	414

Thermochemical equation for the formation of CH₄,

= [716 + 2 × 436.4] – [4 × 414] 

= [716+ 872.8] – [1656] 

= 1588.8 – 1656

= -67.2 kJ mol^-1

∴ Standard enthalpy of formation of CH₄ =