0279 0290 0311 0331 0357 0372 3 6 pts The following equation

0.279 0.290 0.311 0.331 0.357 0.372 **3. (6 pts) The following equation is balanced. H2(g) I2(s)2 HI(g) For the following conditions at 298.15 K, circle G (in kJ). 0.221 atm H2 4.90 6.18 0.440 mol I2 5.18 6.33 0.937 atm HI 5.36 6.54 4.57 5.71 4.78 5.57 5.92 6.70

Solution

Sol:-

We know that relationship between K(equilibrium constant) and gibbs free energy i.e G is:

G = - 2.303RT log K ---(1)

R = universal gas constant = 8.314 × 10^-3 KJ /K mol

Calculation of K from given equation:-

K = [ HI] / [H2] [ I2]

Here , [ HI] = partial pressure of HI

And [ H2] = partial pressure of H2

I.e equilibrium constant is the , ratio of product of molar conc. Of products to the product of molar conc. Of reactants , raised to power their stoichiometric coefficient.

k = (0.937)^2/(0.221)(0.440)

= 0.8779/0.09724

=9.028

Substitute this value in eq.(1)  

G = -2.303× 0.008314× 298.15 log 9.028

= - 5.45 KJ