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NEET CHEMISTRYThermodynamicsMedium

Question

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 mol1^{-1}

B

(xy)(x - y) kJ mol1^{-1}

C

(yx)(y - x) kJ mol1^{-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=ΔH1ΔH2=xy\Delta H_t = \Delta H_1 - \Delta H_2 = x - y

Thus, the heat of transition is (xy)(x - y) kJ mol1^{-1}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicstransitiongraphitediamondfollowingthermochemical

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