Question

For the hypothetical equilibrium A (double arrow) B, you obtain the following data.

T(K)	Keq
260	6.2*10^-2
280	1.6*10^-2
300	5.2*10^-3
320	1.9*10^-3

Determine the values of ∆Go, ∆Ho, and ∆So. What would be the equilibrium constant at T = 325 K?

Answer: 13 kJ/mol, -40 kJ/mol, -177 J/mol•K, 1.5 x 10-3 ( The answer is given I just want to know how they got this answers ) , plz help..

Answer 1

1) we know that

∆Go = -RT lnKeq

R = 8.314

= Temperature = 298 K (approx 300K) Standard temperature

keq = 0.0052

lnKeq = -5.26

∆Go = -8.314 X 300 X (-5.26) = 13119 Joules

∆Go = 13.11 KJ / mole or 13 KJ / mole

We can obtain value of Delta H and Delta S by plotting graph between lnKeq and 1/T (vant hoff equation)

T(K)	Keq	1/T	lnKeq
260	0.062	0.003846	-2.78062
280	0.016	0.003571	-4.13517
300	0.0052	0.003333	-5.2591
320	0.0019	0.003125	-6.2659

The slope = -Delta H / R

from graph slope = 4823.5 = -DeltaH / R

Delta H = -4823.5 X 8.314 = 40102.57 Joules = 40 KJ / mole

Interception = delta S / R

So Delta S = Intercept X R

From graph

Intecept = -21.343

therefore Delta S = -21.343 X 8.314 = 177.44 Joules / mol K

The answers are exactly the same as given

we can put the values in the given equation to obtain value of K

lnK = 4823.5 (1/T) - 21.343

T = 325 so 1/T = 0.00307

lnK = -6.534

So K = antilog (-6.534) = 0.00145 = 1.45 X 10^-3 = 1.5 X 10^-3