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Thermodynamics question

2021 · 25 Jul · Shift 1 · Q21
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Thermodynamics question

2021 · 25 Jul · Shift 1 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol −-− 1 and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ mol −-− 1. The enthalpy of sublimation of the substance (X) in kJ mol −-− 1 is ‾\underline{\hspace{2cm}}​. (in nearest integer)
Numerical answer
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Correct answer: 101

  1. We use Hess's law for phase changes:

ΔHsub=ΔHfus+ΔHvap\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}ΔHsub​=ΔHfus​+ΔHvap​

because sublimation can be considered as:

solid→liquid→vapour\text{solid} \to \text{liquid} \to \text{vapour}solid→liquid→vapour

  1. Given:

ΔHfus=2.8 kJ mol−1\Delta H_{\text{fus}} = 2.8\ \text{kJ mol}^{-1}ΔHfus​=2.8 kJ mol−1 ΔHvap=98.2 kJ mol−1\Delta H_{\text{vap}} = 98.2\ \text{kJ mol}^{-1}ΔHvap​=98.2 kJ mol−1

  1. Therefore,

ΔHsub=2.8+98.2=101.0 kJ mol−1\Delta H_{\text{sub}} = 2.8 + 98.2 = 101.0\ \text{kJ mol}^{-1}ΔHsub​=2.8+98.2=101.0 kJ mol−1

  1. Nearest integer:

101101101

So, the enthalpy of sublimation is 101 kJ mol−1101\ \text{kJ mol}^{-1}101 kJ mol−1.

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