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When 0.0430 mol of HI is heated to 500 K in a 5.00 L sealed container,...

When 0.0430 mol of HI is heated to 500 K in a 5.00 L sealed container, the resulting equilibrium mixture contains 2.85 g of HI. Calculate Kc and Kp for the following decomposition reaction. 2 HI (g) H2 (g) + I2 (g)

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Answer #1

2 HI (g)   H2 (g) + I2 (g).

Kc = [HI]2    HI moles = 0.0430, the volume = 5L

HI concentration (Molarity) = Moles/ litre

= 0.0430/5

= 0.0086M.

Kc = [0.0086]2

Kc = 7.396 x 10-5.

Kp = Kc[RT]n   R = Ideal gass constant = 8.314 jk-1mol-1.

   n = Gaseous product mole no - gaseous reactant mol no.

n = 2-2

n = 0

Kp = Kc[RT]n T = 500 K (according to the question).

Kp = 7.396 x 10-5[8.314x500]0

Kp = 7.396 x 10-5.

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