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Consider the reaction: X2(g)→2X(g). When a vessel initially containing 714 torr of X2 comes to equilibrium...

Consider the reaction: X2(g)→2X(g). When a vessel initially containing 714 torr of X2 comes to equilibrium at 298 K, the equilibrium partial pressure of X is 116 torr . The same reaction is repeated with an initial partial pressure of 777 torr of X2 at 707 K ; the equilibrium partial pressure of X is 561 torr

Find delta H for the reaction (in kJ/mol)

Solutions

Expert Solution

                      X2(g) ---- 2X(g)

Initial              714         0

Final             (714-x)      2x

2x = 116 torr

x = 58 torr

X2 = 714 - 58 = 656 torr

Kp1 = [X]^2/[X2] = (116)^2/656 = 20.512

                      X2(g) ---- 2X(g)

Initial              777         0

Final             (777-x)      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

queen_honey_blossom answered 2 years ago
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