99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
The enthalpies of combustion of C(graphite) and C(diamond) are -393.8 and -395.3 ~kJ ~mol ⁻¹ respectively. The enthalpy of conversion of C (graphite) to C (diamond) is
Options
- A-12.5 ~kJ ~mol ⁻¹
- B-789.1 ~kJ ~mol ⁻¹
- C79.0 ~kJ ~mol ⁻¹
- D1.5 ~kJ ~mol ⁻¹
Correct answer
D. 1.5 ~kJ ~mol ⁻¹
Step-by-step solution
aligned & eq ^ n 1 . . C ( graphite )+ O ₂( ~g ) CO ₂( ~g ) H ₁=-393.8 KJ mol & eq ^ n 2 . . C ( diamond )+ O ₂( ~g ) CO ₂( ~g ) H ₂=-395.3 KJ mol & eq ^ n 1- eq ^ n 2 = H ₁- H ₂ & =-393.8-(-395.3) & =1.5 KJ mol aligned