JEE Main2013ChemistryThermodynamics (C)Actual
Given: (I) aligned & H ₂( g )+ 1 2 O ₂( g ) H ₂ O (l) ; & H ^ _ 298 ~K =-285.9 ~kJ ~mol ⁻¹ aligned (II) aligned & H ₂( g )+ 1 2 O ₂( g ) H ₂ O ( g ); & H ^ _ 298 ~K =-241.8 ~kJ ~mol ⁻¹ aligned The molar enthalpy of vapourisation of water will be :
Options
- A241.8 ~kJ ~mol ⁻¹
- B22.0 ~kJ ~mol ⁻¹
- C44.1 ~kJ ~mol ⁻¹
- D527.7 ~kJ ~mol ⁻¹
Correct answer
A. 241.8 ~kJ ~mol ⁻¹
Step-by-step solution
Given H ₂( g )+ 1 2 O ₂(g) H ₂ O (l) ; H ^ =-285.9 ~kJ ~mol ⁻¹ (1) H ₂( g )+ 1 2 O ₂(g) H ₂ O (g) ; H ^ =-241.8 ~kJ ~mol ⁻¹ (2) We have to calculate H ₂ O (l) H ₂ O ( g ) ; H ^ = ? On substracting eqn. (2) from eqn. (1) we get H ₂ O (l) H ₂ O ( g ) ; H ^ =-241.8-(-285.9)=44.1 ~kJ ~mol ⁻¹