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Q.Calculate the enthalpy change (ΔH) of formation of methane with the help of the following data: C (s) + O2

(g) -> CO2 (g), ΔH = -126 KCal ... (i); 2H2
(g) + O2
(g) -> 2H2O (g), ΔH = -186 KCal ... (ii); CH4
(g) + 2O2
(g) -> CO2
(g) + 2H2O (g), ΔH = -212 KCal ...
(iii) OR Calculate ΔG° for the transformation of oxygen to ozone, (3/2)O2
(g) -> O3 (g), at 298K, if Kp for this transformation is 2.47x10^-29.
Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2024Subjective· 3mImportance★★★★★
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By Hess's Law, adding reaction (i) + (ii) and subtracting (iii) converts C(s) + 2H2(g) into CH4(g), giving an enthalpy of formation of -100 KCal for methane.

Target reaction: C (s) + 2H2 (g) -> CH4 (g), ΔH_f = ?

Given:

  1. C (s) + O2 (g) -> CO2 (g), ΔH1 = -126 KCal
  2. 2H2 (g) + O2 (g) -> 2H2O (g), ΔH2 = -186 KCal
  3. CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g), ΔH3 = -212 KCal Add (i) + (ii): C + 2H2 + 2O2 -> CO2 + 2H2O, ΔH = -126 + (-186) = -312 KCal. Reverse (iii): CO2 + 2H2O -> CH4 + 2O2, ΔH = +212 KCal. Adding these two: C + 2H2 + 2O2 + CO2 + 2H2O -> CO2 + 2H2O + CH4 + 2O2. Cancelling CO2, 2H2O, and 2O2 from both sides leaves: C + 2H2 -> CH4. ΔH_f(CH4) = ΔH1 + ΔH2 - ΔH3 = (-126) + (-186) - (-212) = -312 + 212 = -100 KCal. …

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