Answer to Question #223074 in General Chemistry for 000

Question

. From the following enthalpy changes, C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ H2(g) + ½ O2(g)  H2O(l) ΔH = -285.8 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890.3 kJ calculate the value of ∆Hf for CH4.

Answer 1

We aim at : $C (s) + 2 H2 (g)→ CH4 (g) ; ΔH° = ?$

Multiplying eqn. (ii) with 2, adding to eqn. (i) and then subtracting eqn. (iii) from the sum, i.e., operating eqn. $(i) + 2 × eqn. (ii) – eqn. (iii),$ we get

$C (s) + 2H2 (g) – CH4 (g)$

$ΔHf= – 393.5 + 2 (–285.8) – (–890.3) =74.8$

Hence, enthalpy of formation of methane is : ΔHf= – 74.8 kJ mol