Question

Calculate the enthalpy of formation of CH4 from the following data:

i) C(s) + O2(g) → CO2(g); ∆H = -393.7 kJ/mol

ii) H2(g) + ½ O2(g) → H2O(l); ∆H = -285.8 kJ/mol

iii) CH4(g) + 2 O2(g)→ CO2(g) + 2H2O(l); ∆H = -890.4 kJ/mol​

Answer 1

Answer:

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) → 0; ΔrH° = – 393.5 + 2 (–285.8) – (–890.3) – 74.8 kJ mol–1 or C (s) + 2H2 (g) CH4 (g); ΔrH° = – 74.8 kJ mol–1 Hence, enthalpy of formation of methane is : ΔrH°= – 74.8 kJ mol–1 Read more on Sarthaks.com - https://www.sarthaks.com/520379/calculate-enthalpy-of-formation-of-methane-ch4-from-the-following-data