99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
Given: C graphite + O 2 g → CO 2 g ; Δ r H o = − 393.5 kJ mol − 1 H 2 g + 1 2 O 2 g → H 2 O l ; Δ r H o = - 285.8 k J m o l - 1 C O 2 g + 2 H 2 O l → C H 4 g + 2 O 2 g ; Δ r H o = + 890.3 k J m o l - 1 Based on the above thermochemical equations, the value of Δ r H o at 298 K for the reaction C g r a p h i t e + 2 H 2 g → C H 4 g will be:
Options
- A+ 144.0 k J m o l - 1
- B- 74.8 k J m o l - 1
- C- 144.0 k J m o l - 1
- D+ 74.8 k J m o l - 1
Correct answer
B. - 74.8 k J m o l - 1
Step-by-step solution
For reaction CO 2 (g)+2H 2 O(l) → CH 4 (g)+2O 2 (g) Δ r H o = Σ Δ f H o p r o d u c t s - Σ Δ f H o r e a c t a n t s = Δ r H o C H 4 + 2 × 0 - Δ f H o C O 2 + 2 Δ f H o H 2 O + 890.3 = Δ f H o C H 4 - - 393.5 + 2 × - 285.8 Δ f H o of CH 4 (g) = − 74.8 kJ / mol