Question

for the reaction X2O4 (l) - 2 XO2(g). dU = 2.1 kcal, dS= 20 cal K -1 at 300 K, hence dGis

Answer 1

H=2100+1200

=3300

G=3300-6000

=-2700cal

=-2.7Kcal

Answer 2

Answer: $\Delta G$= - 2700 cal

Explanation:

Relation of $\Delta H$ with $\Delta U$ is given by the formula:

$\Delta H=\Delta E+{\Delta ng}RT$

Where,

$\Delta H$ = enthalpy change = ?

$\Delta E$ = internal energy change= 2.1 kcal = 2100 cal

R = Gas constant = $2calmol^{-1}K^{-1}$

T = temperature = $300K$

$X_2O_4(l)\rightarrow 2XO_2(g)$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-0=2$

Putting values in above equation, we get:

$\Delta H=2100+2\times 2\times 300$

$\Delta H=3300cal$  

According to Gibb's equation:

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

$\Delta G$ = Gibbs free energy  =?

$\Delta H$ = enthalpy change  = 3300 cal

$\Delta S$ = entropy change = 20 cal/K

T = temperature in Kelvin  = 300 K

$\Delta G=3300cal-300(20)=-2700cal$  

Hence, $\Delta G$ is -2700 cal.