Highly selective Backlog Qs for JEE MainChemistryThermodynamics (C)
The combustion of benzene l gives CO 2 g and H 2 O l . Given that heat of combustion of benzene at constant volume is - 3263 .9 kJ mol - 1 at 25 ° C ; the heat of combustion in kJ mol - 1 of benzene at constant pressure will be R = 8 .314 JK - 1 mol - 1
Options
- A- 3267.6
- B4152.6
- C- 452.46
- D3260
Correct answer
A. - 3267.6
Step-by-step solution
C 6 H 6 l + 15 2 O 2 g → 6 CO 2 g + 3 H 2 O ( I ) ∴ ∆ n g = - 3 2 Use, ΔH = ΔU + Δn g RT = - 3263.5 × 10 3 - 3 2 × 8.314 × 298 × 10 - 3 = - 3267 .6 kJ mol - 1