Answer to Question #223074 in General Chemistry for 000

Question #223074

. 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.


1
Expert's answer
2021-08-04T07:13:33-0400

We aim at : "C (s) + 2 H2 (g)\u2192 CH4 (g) ; \u0394H\u00b0 = ?"


Multiplying eqn. (ii) with 2, adding to eqn. (i) and then subtracting eqn. (iii) from the sum, i.e., operating eqn. "(i) + 2 \u00d7 eqn. (ii) \u2013 eqn. (iii)," we get

"C (s) + 2H2 (g) \u2013 CH4 (g)"


"\u0394Hf= \u2013 393.5 + 2 (\u2013285.8) \u2013 (\u2013890.3) =74.8"

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


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