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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 4 April · Evening Shift · Chemistry · Thermodynamics (C)

Consider the given data : (a) HCl(g)+10 H_2 O(l) arrow HCl.10 H_2 O...

Consider the given data :
(a) HCl(g)+10H2O(l)→HCl.10H2O\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O}
ΔH=−69.01 kJ mol−1\Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}
(b) HCl(g)+40H2O(l)→HCl.40H2O\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O}
ΔH=−72.79 kJ mol−1\Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}
Choose the correct statement :

Options

  1. ADissolution of gas in water is an endothermic process
  2. BThe heat of solution depends on the amount of solvent.
  3. CThe heat of dilution for the HCl(HCl⋅10H2O\mathrm{HCl}\left(\mathrm{HCl} \cdot 10 \mathrm{H}_2 \mathrm{O}\right. to \left.\mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O}\right)isis3.78 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
  4. DThe heat of formation of HCl solution is represented by both (a) and (b)

Correct answer

B. The heat of solution depends on the amount of solvent.

Step-by-step solution

From the given information
ΔH\Delta \mathrm{H} is negative so it means dissolution of gas HCl(g)\mathrm{HCl}(\mathrm{g}) is exothermic.
HCl(g)+10H2O(l)→HCl.10H2O\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O} ...(1)
ΔH1=−69.01 kJ mol\Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}}
HCl(g)+40H2O(l)→HCl.40H2O\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O} ...(2)
ΔH2=−72.79 kJ mol\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}
Hence heat of solution depends upon amount of solvent
By equation..(2) - equation
HCl.10H2O+30H2O(ℓ)→HCl⋅40H2O\mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} \cdot 40 \mathrm{H}_2 \mathrm{O}
So Heat of dilution =−72.79−(−69.01)=-72.79-(-69.01)
=−3.78 kJ mol=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}
Hence option (3) is incorrect.
For heat of formation reactant should be in elemental form hence option (4) is incorrect
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