Question

For vaporization of water at 1 atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are 40.63 kJmol⁻¹ and 108.8 JK⁻¹ mol⁻¹, respectively. The temperature when Gibbs energy change ($\Delta G$) for this transformation will be zero, is:
(a) 293.4 K
(b) 273.4 K
(c) 393.4 K
(d) 373.4 K.

Answer 1

Answer:

∆G = 0

∆H - T (∆s) = 0

∆H = T ∆s

T = (40.63/108.8) × 1000 = 373.43 K

So , option (D) is correct

Answer 2

The temperature when Gibbs energy change () for this transformation will be zero is 373.4 K

Explanation:

According to Gibb's equation:

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

$\Delta G$ = Gibbs free energy  = 0

$\Delta H$ = enthalpy change  = $40.63 kJ/mol$= 40630 J/mol

$\Delta S$ = entropy change  =$108.8JK^{-1}mol^{-1}$

T = temperature in Kelvin

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

$0=\Delta H-T\Delta S$

$\Delta H=T\Delta S$

$40630J/mol=T\times 108.8JK^{-1}mol^{-1}[$

$T=373.4K$

Learn More about Gibbs equation

https://brainly.com/question/14121461

https://brainly.com/question/13197203