Chemistry - Thermodynamics - II Question with Solution | TestHub

To find $\Delta G$ for the liquid phase reaction, first calculate $\Delta H$ for the liquid phase.

Given:

1. $4X(g)\to X_4(g)$, $\Delta H_1=-100$ kJ/mol

2. $X(l)\to X(g)$, $\Delta H_{vap,X}=30$ kJ/mol

3. $X_4(l)\to X_4(g)$, $\Delta H_{vap,X_4}=72$ kJ/mol

We want to find $\Delta H_2$ for $4X(l)\to X_4(l)$.

This can be achieved by combining the given reactions:

$4X(l)\to 4X(g)$ ($4\times \Delta H_{vap,X}$)

$4X(g)\to X_4(g)$ ($\Delta H_1$)

$X_4(g)\to X_4(l)$ ($-\Delta H_{vap,X_4}$)

$\Delta H_2=(4\times \Delta H_{vap,X})+\Delta H_1-\Delta H_{vap,X_4}$

$\Delta H_2=(4\times 30kJ/mol)+(-100kJ/mol)-(72kJ/mol)$

$\Delta H_2=120kJ/mol-100kJ/mol-72kJ/mol$

$\Delta H_2=-52kJ/mol$

Now, calculate $\Delta G$ for the liquid phase reaction using the formula:

$\Delta G=\Delta H-T\Delta S$

Given $T=300$ K and $\Delta S=-125$ J/Kmol $=-0.125$ kJ/Kmol.

$\Delta G=-52kJ/mol-(300K\times -0.125kJ/Kmol)$

$\Delta G=-52kJ/mol+37.5kJ/mol$

$\Delta G=-14.5kJ/mol$

The final answer is $\boxed{-14.5kJ/mol}$.