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Thermodynamics question

2023 · 15 Apr · Shift 1 · Q20
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Thermodynamics question

2023 · 15 Apr · Shift 1 · Q20

JEE MainChemistryThermodynamicsNumerical+4 / −1
30.4 kJ30.4 \mathrm{~kJ}30.4 kJ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is 28.4 J K−1 mol−128.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}28.4 J K−1 mol−1 at 1 atm. The melting point of sodium chloride is ‾\underline{\hspace{2cm}}​ K (Nearest Integer)
Numerical answer
View written solutionFree

Correct answer: 1070

  1. Use the relation at the melting point

At the melting point, solid and liquid are in equilibrium, so for the phase change:

ΔG=0=ΔH−TmΔS\Delta G = 0 = \Delta H - T_m \Delta SΔG=0=ΔH−Tm​ΔS

Hence,

Tm=ΔHfusΔSfusT_m = \frac{\Delta H_{fus}}{\Delta S_{fus}}Tm​=ΔSfus​ΔHfus​​

  1. Substitute the given values

Given:

ΔHfus=30.4 kJ mol−1=30.4×103 J mol−1\Delta H_{fus} = 30.4\,\text{kJ mol}^{-1} = 30.4 \times 10^3\,\text{J mol}^{-1}ΔHfus​=30.4kJ mol−1=30.4×103J mol−1

ΔSfus=28.4 J K−1mol−1\Delta S_{fus} = 28.4\,\text{J K}^{-1}\text{mol}^{-1}ΔSfus​=28.4J K−1mol−1

So,

Tm=30.4×10328.4T_m = \frac{30.4 \times 10^3}{28.4}Tm​=28.430.4×103​

  1. Calculate

Tm≈1070.42 KT_m \approx 1070.42\,\text{K}Tm​≈1070.42K

Nearest integer:

1070 K\boxed{1070\,\text{K}}1070K​

  1. Comparison with stored answer

Stored correct answer = 107010701070

My derived answer = 107010701070

So, the derived answer agrees with the stored correct answer.

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