Back to Directory
NEET CHEMISTRYMedium

The heat of transition (ΔHt\Delta H_t) of graphite into diamond would be, given the following thermochemical equations:

  1. C(graphite) + O2_2(g) →\rightarrow CO2_2(g); ΔH=x\Delta H = x kJ
  2. C(diamond) + O2_2(g) →\rightarrow CO2_2(g); ΔH=y\Delta H = y kJ
A

(x+y)(x + y) kJ mol−1^{-1}

B

(x−y)(x - y) kJ mol−1^{-1}

C

(y−x)(y - x) kJ mol−1^{-1}

D

None of these

Step-by-Step Solution

To determine the heat of transition for the reaction C(graphite) →\rightarrow C(diamond), we apply Hess’s Law of Constant Heat Summation. This law states that the total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps .

We are given two combustion reactions: (i) C(graphite) + O2_2(g) →\rightarrow CO2_2(g); ΔH1=x\Delta H_1 = x kJ (ii) C(diamond) + O2_2(g) →\rightarrow CO2_2(g); ΔH2=y\Delta H_2 = y kJ

To find the enthalpy change for the transition from graphite to diamond, we subtract the second equation from the first: [C(graphite) + O2_2(g)] - [C(diamond) + O2_2(g)] →\rightarrow [CO2_2(g)] - [CO2_2(g)] C(graphite) - C(diamond) →\rightarrow 0 C(graphite) →\rightarrow C(diamond)

The enthalpy change for this process (ΔHt\Delta H_t) is calculated by performing the same operation on the enthalpy values: ΔHt=ΔH1−ΔH2=x−y\Delta H_t = \Delta H_1 - \Delta H_2 = x - y

Thus, the heat of transition is (x−y)(x - y) kJ mol−1^{-1}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started