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Consider the following reaction: 2 HI(g) H2(g) + I2(g) If 2.29 moles of HI, 0.309 moles...

Consider the following reaction: 2 HI(g) H2(g) + I2(g) If 2.29 moles of HI, 0.309 moles of H2, and 0.363 moles of I2 are at equilibrium in a 17.8 L container at 774 K, the value of the equilibrium constant, Kp, is

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Answer #1

molarity = moles / volume

[HI] = 2.29 / 17.8 = 0.129 M

[H2] = 0.309 / 17.8 = 0.0174 M

[I2] = 0.363 / 17.8 = 0.0204 M

2 HI(g) --------------------> H2(g) +    I2(g)

0.129                                0.0174       0.0204   ------------ at equilibrium

Kc = [H2][I2] / [HI]^2

      = 0.0174 x 0.0204 / 0.129^2

      = 0.0213

Kp = Kc (RT)\Deltan

     = 0.0213 (0.0821 x 774)^0

     = 0.0213

Kp = 0.0213

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