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Chemistry Question 53 – JEE-MAIN 2025

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Total enthalpy change for freezing of 1 mol of water at 10∘C to ice at −10∘C is _______. \ (Given: ΔfusH=x kJ/mol \ Cp[H2O(l)]=y J mol−1 K−1 \ Cp[H2O(s)]=z J mol−1 K−1

The process involves cooling liquid water, freezing it, and then cooling the resulting ice.

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Ninja Strategy•Sign and Unit Analysis

First, determine the expected sign of the total enthalpy change (exothermic processes yield negative enthalpy). Then, check for unit consistency in the options, ensuring all terms are in the same unit (e.g., Joules) before summation.

Step 1: Break Down the Process✦ Active

The total enthalpy change can be broken down into three sequential steps for 1 mole of water:

1. Cooling liquid water from 10∘C to 0∘C 2. Freezing water at 0∘C to ice at 0∘C 3. Cooling ice from 0∘C to −10∘C
Step 2: Calculate Enthalpy Change for Each Step○ Expand

Given n=1 mol, ΔfusH=x kJ/mol, Cp[H2O(l)]=y J mol−1 K−1, Cp[H2O(s)]=z J mol−1 K−1. We will convert all values to Joules for consistency.

ΔH1=nCp[H2O(l)]ΔT1=(1 mol)×(y J mol−1 K−1)×(0−10) K=−10y J ΔH2=n(−ΔfusH)=(1 mol)×(−x kJ mol−1)=−x kJ=−1000x J ΔH3=nCp[H2O(s)]ΔT2=(1 mol)×(z J mol−1 K−1)×(−10−0) K=−10z J
💡 Teacher's Secret Hint

Remember that freezing is an exothermic process, so the enthalpy change for freezing is negative of the enthalpy of fusion.

Step 3: Sum Total Enthalpy Change○ Expand

The total enthalpy change is the sum of the enthalpy changes for each step:

ΔHtotal=ΔH1+ΔH2+ΔH3 ΔHtotal=−10y J−1000x J−10z J ΔHtotal=−(1000x+10y+10z) J ΔHtotal=−10(100x+y+z) J

Comparing this with the given options, option 2 matches the calculated total enthalpy change.

💡 Teacher's Secret Hint

Always double-check the units in the final expression to ensure consistency with the options provided.

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