Logo
Jeeify
MCQEasyJEE 2025Internal Energy & Enthalpy

JEE Chemistry 2025 Question with Solution

Consider the given data:

(a)  HCl(g)+10H2O(l)→HCl.10 H2OΔH=−69.01 kJ/mol−1(a)\; \text{HCl(g)} + 10\text{H}_2\text{O(l)} \rightarrow \text{HCl}.10\text{ H}_2\text{O} \quad \Delta H = -69.01 \, \text{kJ/mol}^{-1} (b)  HCl(g)+40H2O(l)→HCl.40 H2OΔH=−72.79 kJ/mol−1(b)\; \text{HCl(g)} + 40\text{H}_2\text{O(l)} \rightarrow \text{HCl}.40\text{ H}_2\text{O} \quad \Delta H = -72.79 \, \text{kJ/mol}^{-1}

Choose the correct statement:

  • A

    Dissolution of gas in water is an endothermic process

  • B

    The heat of solution depends on the amount of solvent

  • C

    The heat of dilution for the HCl (HCl.10H2_2O to HCl.40H2_2O) is 3.78 kJ/mol3.78 \, \text{kJ/mol}

  • D

    The heat of formation of HCl solution is represented by both (a) and (b)

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two dissolution processes of HCl(g) in different amounts of water are given with enthalpy changes:

ΔH1=−69.01 kJ/mol\Delta H_1 = -69.01 \, \text{kJ/mol}

for 10H2O10\text{H}_2\text{O}, and

ΔH2=−72.79 kJ/mol\Delta H_2 = -72.79 \, \text{kJ/mol}

for 40H2O40\text{H}_2\text{O}.

Find: Which statement is correct.

Both given values of ΔH\Delta H are negative, so dissolution of HCl(g) in water is exothermic. Therefore, the statement saying dissolution is endothermic is incorrect.

The data show that the enthalpy change is not the same for 1010 and 4040 moles of water. Hence, the heat of solution depends on the amount of solvent.

From the solution working, the difference is calculated as:

ΔHdifference=−72.79 kJ/mol−(−69.01 kJ/mol)=−3.78 kJ/mol\Delta H_{\text{difference}} = -72.79 \, \text{kJ/mol} - (-69.01 \, \text{kJ/mol}) = -3.78 \, \text{kJ/mol}

This supports that the enthalpy changes vary with solvent amount. The solution concludes that statement (2) is the correct one.

Therefore, the correct option is B.

Using the given enthalpy data

Given:

  1. HCl(g)+10H2O(l)→HCl.10 H2O\text{HCl(g)} + 10\text{H}_2\text{O(l)} \rightarrow \text{HCl}.10\text{ H}_2\text{O} with
ΔH=−69.01 kJ/mol\Delta H = -69.01 \, \text{kJ/mol}
  1. HCl(g)+40H2O(l)→HCl.40 H2O\text{HCl(g)} + 40\text{H}_2\text{O(l)} \rightarrow \text{HCl}.40\text{ H}_2\text{O} with
ΔH=−72.79 kJ/mol\Delta H = -72.79 \, \text{kJ/mol}

Find: The correct statement among the four options.

Step 1: Negative enthalpy change means heat is released. So the dissolution process is exothermic, not endothermic. Hence option A is false.

Step 2: Compare the two heats of solution. Since

−69.01≠−72.79-69.01 \neq -72.79

it follows that the heat of solution changes when the amount of water changes from 1010 to 4040 moles. Hence option B is true.

Step 3: the solution computes the change as:

−72.79−(−69.01)=−3.78 kJ/mol-72.79 - (-69.01) = -3.78 \, \text{kJ/mol}

So the sign is negative in the shown working. Option C states 3.78 kJ/mol3.78 \, \text{kJ/mol} without reflecting that sign, and the second approach explicitly marks this statement incorrect.

Step 4: The reactions represent heat of solution, not heat of formation. Hence option D is false.

Therefore, the correct option is B.

Common mistakes

  • Treating a negative value of ΔH\Delta H as endothermic is incorrect. Negative enthalpy change means heat is released. Use the sign convention carefully: negative = exothermic, positive = endothermic.

  • Assuming the heat of solution is independent of solvent amount is incorrect here. The two given values are different for 1010 and 4040 moles of water, so the solvent amount clearly affects the enthalpy change.

  • Ignoring the sign while calculating the enthalpy difference leads to confusion. The shown subtraction gives

    −72.79−(−69.01)=−3.78 kJ/mol-72.79 - (-69.01) = -3.78 \, \text{kJ/mol}

    so always keep track of signs instead of comparing only magnitudes.

Practice more Internal Energy & Enthalpy questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Internal Energy & Enthalpy

Related questions