A mixture of 0.47 mole of H_2 and 3.59 moles of HCl is heated to 2800DegreeC. Ca

Question

A mixture of 0.47 mole of H_2 and 3.59 moles of HCl is heated to 2800DegreeC. Calculate the equilibrium partial pressures of H_2, Cl_2, and HCl if the total pressure is 2.00 atm. For the reaction H_2(g) + Cl_2 (g) = 2HCl (g) K_p is 193 at 2800DegreeC. When heated at high temperature, iodine vapor dissociates as follows: I_2 (g) = 2I (G) In one experiment, a chemist finds that when 0.054 mole of I_2 was placed in a flask of volume 0.48 L at 587 K, the degree of dissociation (that is, the fraction of I_2 dissociated) was 0.0252. Calculate K_c and K_p for the reaction at this temperature.

Explanation / Answer

15) I2(g) <---> 2I(g)

Initial concentration of I2 = moles/volume of solution in litres = 0.054/0.48 = 0.1125 M, [I] = 0 M

Let at eqb., [I2] = (0.1125-x) M & [I] = 2x M

given degree of dissociation = 0.0252

Thus, x = 0.0252*0.1125 = 0.00284 M

thus, Kc = [I]2/[I2] = (0.0057)2/(0.11) = 2.954*10-4

Kp = Kc*(RT)n ; where n = moles of gaseous products - moles of gaseous reactants = 1

Thus, Kp = 1.441

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