IIT-JEE 2007 · Paper 2 · Chemistry Q30

ChemistrySingle correct+3 / -1Question 30 of 66

For the process $\ce{H2O}(l)$ (1 bar, 373 K) → $\ce{H2O}(g)$ (1 bar, 373 K), the correct set of thermodynamic parameters is

\Delta G=0

\Delta S=+\mathrm{ve}

\Delta G=0

\Delta S=-\mathrm{ve}

\Delta G=+\mathrm{ve}

\Delta S=0

\Delta G=-\mathrm{ve}

\Delta S=+\mathrm{ve}

Answer and worked solution

Official answer: A

At the normal boiling point (373 K, 1 bar) liquid and vapour are in equilibrium: $\Delta G=0$ .

Entropy. Vaporisation increases disorder: $\Delta S=\dfrac{\Delta H_\text{vap}}T>0$ .

(A)

Topics:Gibbs energy and spontaneity Entropy

IIT-JEE 2007 · Paper 2, Chemistry question 30