For vaporisation of water at 1 ~atm pressure, the values of Δ H and Δ S are 40.63 ~kJ ~mol^ -1 and 108.8 JK^ -1 ~mol^ -1 , respectively. The temperature when Gibbs energy change (Δ G) for this transformation will be

Correct answer: C. 373.4 ~K aligned Δ G& =Δ H-T Δ S Δ G& =0( given ) Δ H& =T Δ S, T& =(40.63 × 10^3)/(108.8)=373.4 K aligned

NEET 2010

NEET 2010 · Chemistry · Thermodynamics (C)

For vaporisation of water at 1 ~atm pressure, the values of Δ H and...

For vaporisation of water at

1 \mathrm{~atm}

pressure, the values of

\Delta \mathrm{H}

and

\Delta S

are

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

and

108.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

, respectively. The temperature when Gibbs energy change

(\Delta G)

for this transformation will be zero, is

C. $373.4 \mathrm{~K}$

\begin{aligned} \Delta \mathrm{G}& =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\\ \Delta \mathrm{G}& =0(\text{ given })\\ \Delta \mathrm{H}& =\mathrm{T} \Delta \mathrm{S},\\ \mathrm{T}& =\frac{40.63 \times 10^3}{108.8}=373.4 \mathrm{ K} \end{aligned}