Question

At 0 degree Celsius ice and water ate in equilibrium and ∆H is 6 kJ/mol find ∆S and ∆G

Answer 1

ΔG=0 at equlibrium

ΔH=TΔS

6000/273 = ΔS

ΔS=21.97J/mol

Answer 2

∆S is 0.02 kJ/Kmol and ∆G is 0 kJ

Explanation:

According to Gibb's equation:

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

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change  = 6 kJ/mol

$\Delta S$ = entropy change  = ?

T = temperature in Kelvin  = $0^0C=273K$

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

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

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

For equilibrium reaction:

$H_2O(s)\rightleftharpoons H_2O(l)$ $\Delta G$= 0,

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

$\Delta H=T\Delta S$

$6kJ/mol=273K\times \Delta S$

$\Delta S=0.02kJ/Kmol$

Learn More about Gibbs equation

https://brainly.com/question/13136906

https://brainly.com/question/9748006