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Thermodynamics question

2025 · 4 Apr · Shift 2 · Q15
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Thermodynamics question

2025 · 4 Apr · Shift 2 · Q15

JEE MainChemistryThermodynamicsMCQ+4 / −1
Consider the given data : (a) HCl(g)+10H2O(l)→HCl.10H2OΔH=−69.01 kJ mol−1\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}HCl(g)+10H2​O(l)→HCl.10H2​OΔH=−69.01 kJ mol−1(b) HCl(g)+40H2O(l)→HCl.40H2OΔH=−72.79 kJ mol−1\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}HCl(g)+40H2​O(l)→HCl.40H2​OΔH=−72.79 kJ mol−1 Choose the correct statement :
  1. A
    The heat of dilution for the HCl(HCl.10H2O\mathrm{HCl}\left(\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\right.HCl(HCl.10H2​O to HCl.40H2O)\left.\mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\right)HCl.40H2​O) is 3.78 kJ mol−13.78 \mathrm{~kJ} \mathrm{~mol}^{-1}3.78 kJ mol−1.
  2. B
    Dissolution of gas in water is an endothermic process.
  3. C
    The heat of solution depends on the amount of solvent.
  4. D
    The heat of formation of HCl solution is represented by both (a) and (b).
View written solutionFree

Correct answer: C

  1. Interpret the given reactions

The data given are:

(a)HCl(g)+10 H2O(l)→HCl⋅10H2O,ΔH=−69.01 kJ mol−1\text{(a)}\quad \mathrm{HCl}(g) + 10\,\mathrm{H_2O}(l) \rightarrow \mathrm{HCl}\cdot 10\mathrm{H_2O}, \qquad \Delta H = -69.01\,\mathrm{kJ\,mol^{-1}}(a)HCl(g)+10H2​O(l)→HCl⋅10H2​O,ΔH=−69.01kJmol−1 (b)HCl(g)+40 H2O(l)→HCl⋅40H2O,ΔH=−72.79 kJ mol−1\text{(b)}\quad \mathrm{HCl}(g) + 40\,\mathrm{H_2O}(l) \rightarrow \mathrm{HCl}\cdot 40\mathrm{H_2O}, \qquad \Delta H = -72.79\,\mathrm{kJ\,mol^{-1}}(b)HCl(g)+40H2​O(l)→HCl⋅40H2​O,ΔH=−72.79kJmol−1

Both equations represent solution of gaseous HCl in different amounts of water.

Since ΔH\Delta HΔH is negative in both cases, the process is exothermic.

  1. Check option B

Option B says: dissolution of gas in water is an endothermic process.

But here:

ΔH=−69.01 kJ mol−1and−72.79 kJ mol−1\Delta H = -69.01\,\mathrm{kJ\,mol^{-1}} \quad \text{and} \quad -72.79\,\mathrm{kJ\,mol^{-1}}ΔH=−69.01kJmol−1and−72.79kJmol−1

Both are negative, so dissolution is exothermic, not endothermic.

So, B is false.

  1. Find heat of dilution from HCl⋅10H2O\mathrm{HCl}\cdot 10\mathrm{H_2O}HCl⋅10H2​O to HCl⋅40H2O\mathrm{HCl}\cdot 40\mathrm{H_2O}HCl⋅40H2​O

We use Hess's law.

Given:

HCl(g)+10H2O(l)→HCl⋅10H2OΔH1=−69.01\mathrm{HCl}(g) + 10\mathrm{H_2O}(l) \rightarrow \mathrm{HCl}\cdot 10\mathrm{H_2O} \qquad \Delta H_1=-69.01HCl(g)+10H2​O(l)→HCl⋅10H2​OΔH1​=−69.01 HCl(g)+40H2O(l)→HCl⋅40H2OΔH2=−72.79\mathrm{HCl}(g) + 40\mathrm{H_2O}(l) \rightarrow \mathrm{HCl}\cdot 40\mathrm{H_2O} \qquad \Delta H_2=-72.79HCl(g)+40H2​O(l)→HCl⋅40H2​OΔH2​=−72.79

To go from HCl⋅10H2O\mathrm{HCl}\cdot 10\mathrm{H_2O}HCl⋅10H2​O to HCl⋅40H2O\mathrm{HCl}\cdot 40\mathrm{H_2O}HCl⋅40H2​O, add 30 H2O30\,\mathrm{H_2O}30H2​O:

HCl⋅10H2O+30H2O(l)→HCl⋅40H2O\mathrm{HCl}\cdot 10\mathrm{H_2O} + 30\mathrm{H_2O}(l) \rightarrow \mathrm{HCl}\cdot 40\mathrm{H_2O}HCl⋅10H2​O+30H2​O(l)→HCl⋅40H2​O

Its enthalpy change is:

ΔH=ΔH2−ΔH1=−72.79−(−69.01)=−3.78 kJ mol−1\Delta H = \Delta H_2 - \Delta H_1 = -72.79 - (-69.01) = -3.78\,\mathrm{kJ\,mol^{-1}}ΔH=ΔH2​−ΔH1​=−72.79−(−69.01)=−3.78kJmol−1

So the heat of dilution is −3.78 kJ mol−1-3.78\,\mathrm{kJ\,mol^{-1}}−3.78kJmol−1.

Option A says it is +3.78 kJ mol−1+3.78\,\mathrm{kJ\,mol^{-1}}+3.78kJmol−1, without the negative sign. Since the process is exothermic, the correct value should be:

−3.78 kJ mol−1-3.78\,\mathrm{kJ\,mol^{-1}}−3.78kJmol−1

So, A is false.

  1. Check option C

The enthalpy of solution in (a) and (b) is different:

−69.01 kJ mol−1≠−72.79 kJ mol−1-69.01\,\mathrm{kJ\,mol^{-1}} \neq -72.79\,\mathrm{kJ\,mol^{-1}}−69.01kJmol−1=−72.79kJmol−1

The only difference between the two cases is the amount of solvent used.

Hence, the heat of solution depends on the amount of solvent.

So, C is true.

  1. Check option D

Option D says both (a) and (b) represent heat of formation of HCl solution.

This is not the standard meaning of heat of formation. These reactions represent enthalpy of solution (or integral heat of solution) of HCl gas in water at different concentrations, not heat of formation from constituent elements in standard states.

So, D is false.

  1. Final conclusion
  • A: False, because heat of dilution is −3.78 kJ mol−1-3.78\,\mathrm{kJ\,mol^{-1}}−3.78kJmol−1, not +3.78+3.78+3.78.
  • B: False, dissolution here is exothermic.
  • C: True.
  • D: False.

Therefore, the correct option is:

C\boxed{\text{C}}C​
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