NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
The standard enthalpy of formation of gaseous H 2 O at 298 K is − 241 .82 kJ mo l - 1 . Calculate ΔH f o at 373 K, given the following values of the molar heat capacities at constant pressure. Molar heat capacity of H 2 g = 33.58 J K - 1 m o l - 1 Molar heat capacity of H 2 g = 28.84 J K - 1 m o l - 1 Molar heat capacity of O 2 g = 29.37 J K - 1 m o l - 1 Assume that the heat capacities are independent of temperature
Options
- A- 242.6 kJ m o l - 1
- B- 485.2 kJ m o l - 1
- C- 121.3 kJ m o l - 1
- D- 286.4 kJ m o l - 1
Correct answer
A. - 242.6 kJ m o l - 1
Step-by-step solution
The reaction is H 2 g + 1 2 O 2 g → H 2 O g Δ C p o = C p , m o H 2 O , g - C p , m o H 2 , g + 1 2 C p , m o O 2 = 33.58 - 28.84 + 1 2 29.37 = - 9.94 JK - 1 m o l - 1 Using Kirchhoff's equation, Δ H ° 373 K = Δ H ° 298 K + T 2 - T 1 Δ C p o = - 241.82 + 373 - 298 × - 9.94 = - 242.6 kJ mo l - 1