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Thermodynamics question

2017 · 8 Apr · Shift 1 · Q17
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Thermodynamics question

2017 · 8 Apr · Shift 1 · Q17

JEE MainChemistryThermodynamicsMCQ+4 / −1
The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is : (Given Δ\DeltaΔ fusH = 6 kJ mol −-− 1 at 0oC, Cp(H2OH_2OH2​O, ℓ\ellℓ = 75.3J mol −-− 1 K −-− 1) Cp(H2OH_2OH2​O s) =36.8 J mol −-− 1 K −-− 1)
  1. A
    5.44 kJ mol −-− 1
  2. B
    5.81 kJ mol −-− 1
  3. C
    6.56 kJ mol −-− 1
  4. D
    6.00 kJ mol −-− 1
View written solutionFree

Correct answer: C

  1. Break the process into steps

We need the enthalpy change for: H2O(ℓ,5∘C)→H2O(s,−5∘C)\text{H}_2\text{O}(\ell, 5^\circ C) \to \text{H}_2\text{O}(s, -5^\circ C)H2​O(ℓ,5∘C)→H2​O(s,−5∘C)

A convenient path is:

  1. Cool liquid water from 5∘C5^\circ C5∘C to 0∘C0^\circ C0∘C
  2. Freeze water at 0∘C0^\circ C0∘C
  3. Cool ice from 0∘C0^\circ C0∘C to −5∘C-5^\circ C−5∘C
  1. Step 1: Cooling water from 5∘C5^\circ C5∘C to 0∘C0^\circ C0∘C

ΔH1=nCp(ℓ)ΔT\Delta H_1 = n C_p(\ell) \Delta TΔH1​=nCp​(ℓ)ΔT

Here, n=1,Cp(ℓ)=75.3 J mol−1K−1,ΔT=0−5=−5 Kn=1, \quad C_p(\ell)=75.3\,\text{J mol}^{-1}\text{K}^{-1}, \quad \Delta T = 0-5 = -5\,\text{K}n=1,Cp​(ℓ)=75.3J mol−1K−1,ΔT=0−5=−5K

So, ΔH1=1×75.3×(−5)=−376.5 J mol−1\Delta H_1 = 1 \times 75.3 \times (-5) = -376.5\,\text{J mol}^{-1}ΔH1​=1×75.3×(−5)=−376.5J mol−1

  1. Step 2: Freezing at 0∘C0^\circ C0∘C

Given enthalpy of fusion: ΔHfus=+6 kJ mol−1\Delta H_{\text{fus}} = +6\,\text{kJ mol}^{-1}ΔHfus​=+6kJ mol−1

For freezing, the sign reverses: ΔH2=−6 kJ mol−1=−6000 J mol−1\Delta H_2 = -6\,\text{kJ mol}^{-1} = -6000\,\text{J mol}^{-1}ΔH2​=−6kJ mol−1=−6000J mol−1

  1. Step 3: Cooling ice from 0∘C0^\circ C0∘C to −5∘C-5^\circ C−5∘C

ΔH3=nCp(s)ΔT\Delta H_3 = n C_p(s) \Delta TΔH3​=nCp​(s)ΔT

Cp(s)=36.8 J mol−1K−1,ΔT=−5−0=−5 KC_p(s)=36.8\,\text{J mol}^{-1}\text{K}^{-1}, \quad \Delta T = -5 - 0 = -5\,\text{K}Cp​(s)=36.8J mol−1K−1,ΔT=−5−0=−5K

Thus, ΔH3=1×36.8×(−5)=−184 J mol−1\Delta H_3 = 1 \times 36.8 \times (-5) = -184\,\text{J mol}^{-1}ΔH3​=1×36.8×(−5)=−184J mol−1

  1. Total enthalpy change

ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3ΔH=ΔH1​+ΔH2​+ΔH3​

ΔH=−376.5−6000−184=−6560.5 J mol−1\Delta H = -376.5 - 6000 - 184 = -6560.5\,\text{J mol}^{-1}ΔH=−376.5−6000−184=−6560.5J mol−1

ΔH=−6.56 kJ mol−1\Delta H = -6.56\,\text{kJ mol}^{-1}ΔH=−6.56kJ mol−1

  1. Interpretation

Since freezing and cooling release heat, the enthalpy change is negative. The options are given as magnitudes, so the correct option corresponds to: 6.56 kJ mol−16.56\,\text{kJ mol}^{-1}6.56kJ mol−1

So, Option C is correct.

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