JEE Challenger
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ChemHardAdv 2023 • Paper 2

Enthalpy Change in Alpha to Beta Phase Transition

Comprehension Passage

The entropy versus temperature plot for phases α\alpha and β\beta at 1 bar1\text{ bar} pressure is given. STS_{\text{T}} and S0S_0 are entropies of the phases at temperatures T\text{T} and 0 K0\text{ K}, respectively.

The transition temperature for α\alpha to β\beta phase change is 600 K600\text{ K} and Cp,β−Cp,α=1 J mol−1 K−1C_{\text{p},\beta} - C_{\text{p},\alpha} = 1\text{ J mol}^{-1}\text{ K}^{-1}. Assume (Cp,β−Cp,α)(C_{\text{p},\beta} - C_{\text{p},\alpha}) is independent of temperature in the range of 200200 to 700 K700\text{ K}. Cp,αC_{\text{p},\alpha} and Cp,βC_{\text{p},\beta} are heat capacities of α\alpha and β\beta phases, respectively.

The value of enthalpy change, Hβ−HαH_{\beta} - H_{\alpha} (in J mol−1\text{J mol}^{-1}), at 300 K300\text{ K} is _____.

Question Diagram 1
Official Numerical Answer300

Topics & Concepts

ThermodynamicsEnthalpyEntropy and Second LawFirst Law of Thermodynamics

Step-by-Step Solution

To find the value of the enthalpy change, Hβ−HαH_{\beta} - H_{\alpha}, at T=300 KT = 300\text{ K}, we can follow these steps:

Step 1: Calculate the entropy change of phase transition at 600 K600\text{ K}

According to the Third Law of Thermodynamics, for perfectly crystalline substances, the entropy at 0 K0\text{ K} is zero: S0,α=S0,β=0 J mol−1 K−1S_{0,\alpha} = S_{0,\beta} = 0\text{ J mol}^{-1}\text{ K}^{-1}

From the given plot at the phase transition temperature Ttrans=600 KT_{\text{trans}} = 600\text{ K}:

  • For the β\beta phase: S600,β−S0,β=6 J mol−1 K−1  ⟹  Sβ(600 K)=6 J mol−1 K−1S_{600,\beta} - S_{0,\beta} = 6\text{ J mol}^{-1}\text{ K}^{-1} \implies S_{\beta}(600\text{ K}) = 6\text{ J mol}^{-1}\text{ K}^{-1}
  • For the α\alpha phase: S600,α−S0,α=5 J mol−1 K−1  ⟹  Sα(600 K)=5 J mol−1 K−1S_{600,\alpha} - S_{0,\alpha} = 5\text{ J mol}^{-1}\text{ K}^{-1} \implies S_{\alpha}(600\text{ K}) = 5\text{ J mol}^{-1}\text{ K}^{-1}

Thus, the entropy change for the phase transition α→β\alpha \rightarrow \beta at 600 K600\text{ K} is: ΔStrans(600 K)=Sβ(600 K)−Sα(600 K)=6−5=1 J mol−1 K−1\Delta S_{\text{trans}}(600\text{ K}) = S_{\beta}(600\text{ K}) - S_{\alpha}(600\text{ K}) = 6 - 5 = 1\text{ J mol}^{-1}\text{ K}^{-1}

Step 2: Calculate the enthalpy change of phase transition at 600 K600\text{ K}

At the transition temperature Ttrans=600 KT_{\text{trans}} = 600\text{ K}, the two phases are in thermodynamic equilibrium (ΔGtrans=0\Delta G_{\text{trans}} = 0): ΔGtrans(600 K)=ΔHtrans(600 K)−Ttrans⋅ΔStrans(600 K)=0\Delta G_{\text{trans}}(600\text{ K}) = \Delta H_{\text{trans}}(600\text{ K}) - T_{\text{trans}} \cdot \Delta S_{\text{trans}}(600\text{ K}) = 0

ΔHtrans(600 K)=Ttrans⋅ΔStrans(600 K)\Delta H_{\text{trans}}(600\text{ K}) = T_{\text{trans}} \cdot \Delta S_{\text{trans}}(600\text{ K}) ΔHtrans(600 K)=600 K×1 J mol−1 K−1=600 J mol−1\Delta H_{\text{trans}}(600\text{ K}) = 600\text{ K} \times 1\text{ J mol}^{-1}\text{ K}^{-1} = 600\text{ J mol}^{-1}

Step 3: Calculate the enthalpy change at 300 K300\text{ K} using Kirchhoff's Law

Using Kirchhoff's equation for the temperature dependence of the enthalpy change: ΔH(600 K)−ΔH(300 K)=∫300600ΔCp dT\Delta H(600\text{ K}) - \Delta H(300\text{ K}) = \int_{300}^{600} \Delta C_{\text{p}} \, dT

where ΔCp=Cp,β−Cp,α=1 J mol−1 K−1\Delta C_{\text{p}} = C_{\text{p},\beta} - C_{\text{p},\alpha} = 1\text{ J mol}^{-1}\text{ K}^{-1} (given to be independent of temperature).

Integrating gives: ΔH(600 K)−ΔH(300 K)=ΔCp⋅(600−300)\Delta H(600\text{ K}) - \Delta H(300\text{ K}) = \Delta C_{\text{p}} \cdot (600 - 300) 600−ΔH(300 K)=1 J mol−1 K−1×(600−300) K600 - \Delta H(300\text{ K}) = 1\text{ J mol}^{-1}\text{ K}^{-1} \times (600 - 300)\text{ K} 600−ΔH(300 K)=300 J mol−1600 - \Delta H(300\text{ K}) = 300\text{ J mol}^{-1} ΔH(300 K)=600−300=300 J mol−1\Delta H(300\text{ K}) = 600 - 300 = 300\text{ J mol}^{-1}

Thus, the value of the enthalpy change Hβ−HαH_{\beta} - H_{\alpha} at 300 K300\text{ K} is 300.

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Enthalpy Change in Alpha to Beta Phase Transition | Chemistry PYQ Solution - JEE Challenger