NCERT Class XI Chemistry Thermodynamics Solutions

Question : 5

Total: 22

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

- 74.8 kJmol–1
- 52.27 kJmol–1
+ 74.8 kJmol–1
+ 52.26 kJmol–1

Solution:

CH4+2O2 → CO2+2H2O ; ΔH = - 890.3 kJ ... (1)
C + O2 → CO2 ; ΔH = - 393.5 kJ ... (2)
H2+

O2 → H2O ; ΔH = - 285.8 kJ or 2H2+O2 → 2H2O ; ΔH = - 571.6 kJ ... (3)
Add equations (2) and (3) and subtract equation (1) we get, C + 2H2 → CH4
ΔH = –393.5 – 571.6 + 890.3 = –74.8 kJ/mol
∴ Enthalpy of formation of CH4 is –74.8 kJ/mol.

Go to Question:

Prev Question

Next Question