Question

At 1 atm pressure, $\Delta S={75JK}^{-1}{mol}^{-1}$ and $\Delta H=30kJ{mol}^{-1}$.

Find the temperature of vapour, at equilibrium, at 1 atm.

• A. 400 K
• B. 330 K
• C. 200 K
• D. 110 K

Answer 1

The correct option is A 400 K

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

At equilibrium, $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{30\times {10}^3}{75}=400K$

Hence, the correct option is $A$