The enthalpy and entropy change of trimerization of gas $(3 A (g) ⟶ A_{3} (g))$ in gas phase $300$ K at $- 100 kJ / mole$ and $- 400 J / K mole$ respectively. The enthalpy of vapourisation for liquids A and A3 are $25 kJ / mole$ and $50 kJ / mole$ respectively. If the boiling points of A and A3 are $300 K$ and $400 K$ respectively
Given : $C_{p_{m} A_{3} (g)} = C_{p_{m} A_{3} (l)}, d_{A (l)} = 1 .2 g / ml$
$d_{A_{3} (l)} = 1 .5 g / ml$ , molar mass of A = 20 gram
Assume : Density of A₃(l) and A(l) are independent of pressure.

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$ΔH$ for trimerisation in liquid phase at $300 K = - 125 {kJ mol}^{- 1}$

At standard state of $300 K ΔG$ of the reaction $3 A (l) ⟶ A_{3} (l) is 7500 J$

The equilibrium pressure of the reaction:

$3 A (l) ⟶ A_{3} (l)$ is $7 .501 \times 10^{8} Pa$ at $300 K$

$ΔS$ for trimerisation in liquid phase at $300 K = - 275 {kJ mol}^{- 1}$

answer is C, D.

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Detailed Solution

$3 A (g) ⟶ A_{3} (g); ΔH = - 100 kJ / mole$

$ΔS = - 400 kJ / mole$

${ΔH}_{vap} (A, l) = 25 kJ / mole, {ΔS}_{vap} (A_{3}, l) = 50 kJ / mole$

Molar mass of $A = 20 g$

$\begin{array}{l} C_{P_{m},} A_{3} (g) = C_{Pm}, A_{3} (l); d_{A} (l) = 1 .2 g / mL; \\ d_{A_{3} (i)} = 1 .5 g / mL \end{array}$

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