Â
X2(g) ---- 2X(g)
Initial            Â
714Â Â Â Â Â Â Â Â 0
Final           Â
(714-x)Â Â Â Â Â 2x
2x = 116 torr
x = 58 torr
X2 = 714 - 58 = 656 torr
Kp1 = [X]^2/[X2] = (116)^2/656 = 20.512
                    Â
X2(g) ---- 2X(g)
Initial            Â
777 Â Â Â Â Â Â Â 0
Final           Â
(777-x)Â Â Â Â Â 2x
2x = 561 torr
x = 280.5 torr
X2 = 777 - 280.5 = 496.5 torr
Kp2 = [X]^2/[X2] = (561)^2/496.5 = 633.879
Using the claussius clapeyron equation we
get
ln(kp2/kp1) = DeltaH/R * (1/T1 - 1/T2)
ln(633.879/20.512) = Delta H/8.314 * (1/298 -
1/707)
Delta H = 14693.33 J/mol = 14.693 KJ/mol
