JEE Advanced 2023 — Paper 2 — CHEM-IV solution

JEE Advanced2023Paper 2CHEM-IV

PARAGRAPH I

The entropy versus temperature plot for phases $α$ and $β$ at $1$ bar pressure is given. $S_{T}$ and $S_{0}$ are entropies of the phases at temperatures $T$ and $0$ K, respectively.

The transition temperature for $α$ to $β$ phase change is $600$ K and $C_{P, β} - C_{P, α} = 1$ J mol $^{- 1}$ K $^{- 1}$ . Assume $(C_{P, β} - C_{P, α})$ is independent of temperature in the range of $200$ to $700$ K.

Q2. The value of enthalpy change, $H_{β} - H_{α}$ (in J mol $^{- 1}$ ), at $300$ K is __.

Your answer

Solution

At the transition temperature, the two phases are in equilibrium so $Δ G_{t r an s} (600) = 0$ , which gives

$Δ H_{t r an s} (600) = T \cdot Δ S_{t r an s} (600) = 600 \times 1 = 600$ J mol $^{- 1}$ .

Use Kirchhoff's relation for the temperature dependence of $Δ H$ :

$\frac{d ( Δ H )}{d T} = Δ C_{P} = C_{P, β} - C_{P, α} = 1 J mol^{- 1} K^{- 1}$ .

Therefore

$Δ H (600) - Δ H (300) = Δ C_{P} \times (600 - 300) = 1 \times 300 = 300 J mol^{- 1}$ .

$∴ H_{β} - H_{α} at 300 K = 600 - 300 = 300 J mol^{- 1}$ .

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