If the enthalpy change for the transition of liquid water to steam is 30 ~kJ ~mol^ -1 at 27^ C , the entropy change for the process would be

Correct answer: C. 100 ~J ~mol^ -1 ~K^ -1 aligned & ^ ^ = ^ ^ -T ^ ^ & Given, _ vap. =30 ~kJ ~mol^ -1 & ^ ^ =0 at equilibrium, & _ vap = Δ H_ vap T & = 30 × 10^3 ~J ~mol^ -1 300 ~K & =100 ~J ~mol^ -1 ~K^ -1 & aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2011 · Chemistry · Thermodynamics (C)

If the enthalpy change for the transition of liquid water to steam...

If the enthalpy change for the transition of liquid water to steam is

30 \mathrm{~kJ} \mathrm{~mol}^{-1}

27^{\circ} \mathrm{C}

, the entropy change for the process would be

Options

A $1.0 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
B $0.1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
C $100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
D $10 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Correct answer

C. $100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Step-by-step solution

\begin{aligned}& \DeltaG^{{}^{\circ}}=\DeltaH^{{}^{\circ}}-T \DeltaS^{{}^{\circ}} \\ & \text{ Given, } \DeltaH_{\text{vap. }}=30 \mathrm{~kJ} \mathrm{~mol}^{-1}\\ & \DeltaG^{{}^{\circ}}=0 \text{ at equilibrium, }\\ & \DeltaS_{\text{vap }}=\frac{\Delta H_{\text{vap }}}{T}\\ & =\frac{30 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{300 \mathrm{~K}}\\ & =100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\\ & \end{aligned}

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