Reaction of ethene, CHa, with iodine is an endothermic process which is being co

Question

Reaction of ethene, CHa, with iodine is an endothermic process which is being considered as a route t make di-iodoethane accordin g to the following equation: GH4 (g) + 12 (g) C2H412 (g) At 50 °C, the equilibrium partial pressures of C2H4, iodine and C2H412 were found to be PGJ,-107.6 mmHg, P12-1.17 mmHg and PCzHal,-1.14 mmHg. (a) Write an expression for the equilibrium constant, Kp, for this reaction. (b) Determine the value of the equilibrium constant, Kp for this reaction at 50 °C (c) Determine the value of the equilibrium constant Ke for this reaction at 50 °c. (d) Evaluate the equilibrium constant for the related process: C2H4l2 (9) C2H4 )12)

Explanation / Answer

For the given reaction,

Kp = [pC2H4I2]/[pC2H4][pI2]

feeding the given equilibrium values,

Kp = (1.14)/(107.6 x 1.17) = 0.009

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added more iodine,

pI2 = 3 mbar = 2.25 mmHg

So,

F. Qp = [pC2H4I2]/[pC2H4][pI2]

feeding the given equilibrium values,

Kp = (1.14)/(107.6 x 2.25) = 0.005

G. As Qp < Kp, the system will shift to the right, more products would form.

H. ICE chart

           C2H4 +   I2 ----->    C2H4I2

I          107.6      2.25              1.14

C -x -x +x

E       107.6-x   2.25-x          1.14+x

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feeding values into Kp equation,

0.009 = (1.14+x)/(107.6-x)(2.25-x)

0.009x^2 + 2.18 - x = 1.14-x

0.009x^2 + 2x + 1.04 = 0

x = 0.55 mmHg

new equilibrium partial pressures would be,

pC2H4 = 107.6 - 0.55 = 107.05 mmHg

pI2 = 2.25 – 0.55 = 1.7 mmHg

pC2H4I2 = 1.14 – 0.55 = 1.69 mmHg

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