MCQ
The entropy versus temperature plot for phases $\alpha$ and $\beta$ at 1 bar pressure is given. $S_{\mathrm{T}}$ and $S_0$ are entropies of the phases at temperatures $\mathrm{T}$ and $0 \mathrm{~K}$, respectively.

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The transition temperature for $\alpha$ to $\beta$ phase change is $600 \mathrm{~K}$ and $C_{p, \beta}-C_{p, \alpha}=1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. Assume $\left(C_{p, \beta}-C_{p, \alpha}\right)$ is independent of temperature in the range of 200 to $700 \mathrm{~K} . \mathrm{C}_{p, \alpha}$ and $C_{p, \beta}$ are heat capacities of $\alpha$ and $\beta$ phases, respectively.

($1$)The value of entropy change, $\mathrm{S}_\beta-\mathrm{S}_\alpha$ (in $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ), at $300 \mathrm{~K}$ is. . . . . . .

[Use : $\ln 2=0.69$  Given : $S_\beta-S_\alpha=0$ at $\left.0 \mathrm{~K}\right]$

($2$) The value of enthalpy change, $\mathrm{H}_\beta-\mathrm{H}_\alpha$ (in $J$ mol ${ }^{-1}$ ), at $300 \mathrm{~K}$ is

Give the answer quetion ($1$) and ($2$)

  • $31,300$
  • B
    $31,400$
  • C
    $32,500$
  • D
    $34,600$

Answer

Correct option: A.
$31,300$
a
($1$) At $1$ bar

$\alpha \longrightarrow \beta$

$S_{\alpha(600)}^{\circ}=S_{\alpha(300)}^{\circ}+C_{P(\alpha)} \ell n \frac{600}{300}$

$S_{\beta(600)}^{\circ}=S_{\beta(300)}^{\circ}+C_{P(\beta)} \ell n \frac{600}{300}$

$S_{\beta(600)}^{\circ}-S_{\alpha(600)}^{\circ}=S_{\beta(300)}^{\circ}-S_{\alpha(300)}^{\circ}+\left(C_{P(\beta)}-C_{P(\alpha)}\right) \ell \mathrm{n} 2$

$6-5=S_{\beta(300)}^{\circ}-S_{\alpha(300)}^{\circ}+1 \times \ell \mathrm{n} 2$

$1=S_{\beta(300)}^{\circ}-S_{\alpha(300)}^{\circ}+0.69$

So $S_{\beta(300)}^{\circ}-S_{a(300)}^{\circ}=0.31$

($2$) As the phase transition temperature is $600 \mathrm{~K}$

So at $600 \mathrm{~K} \Delta \mathrm{G}_{\mathrm{rn}}^{\circ}=0$

So $\Delta \mathrm{H}^{\circ}{ }_{\text {rasction (600) }}=\mathrm{T} \Delta \mathrm{S}_{\text {raction (600) }}$

$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$

So

$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{p}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$

$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$

$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300 \text { Joule/mole. }$

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