Question

Question 6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be

(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.

Class XI Thermodynamics Page 182

Answer 1

combustion of methane :
CH4 + 2O2--> CO2 + 2H2O; ∆H1 =-890.3 KJ/mol
reverse of this reaction is
CO2 + 2H2O -->CH4 + 2O2;∆H1' = 890.3 KJ/mol------(1)

Combustion of graphite:
C + O2 --> CO2;∆H2=-393.5KJ/mol------(2)

combustion of dihydrogen:
H2 + (1/2)O2---> H2O;∆H3=-285.8 KJ/mol-------(3)

Now, we have to find enthalpy of formation of CH4
e.g., C + 2H2-->CH4:∆Hf =?

equation (2) +2 × equation (3)+equation (1)

C + O2 + 2H2 + O2 + CO2 + 2H2O--> CO2 + 2H2O + CH4 +2O2; ∆Hf = ∆H3 +2∆H2+H1'

So, ∆Hf = (-393.5-2×285.8+ 890.3) KJ/mol
= -74.8 KJ/mol

Hence, option (1) is correct.

Answer 2

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