A 0.682 g sample of ICl(g) is placed in a 765 ml reaction vessel at 682 K. when equilibrium is reached in the reaction 2ICl(g) <=> I2(g) + Cl2(g) 0.0383 g of I2 is found in the mixture. What is Kc for this reaction?

1 Answer
Mar 10, 2018

K_C = color(blue)0.00149

Explanation:

The equilibrium constant of concentration (Kc) expresses the ratio of concentrations [ C ] of products over reactants for a reaction that is at equilibrium.

2 ICl rightleftharpoons "Cl_2 + I_2

The equilibrium constant is written as:

K_C = ([I_2] [Cl_2])/([ICl]^2)

"If :"

K_C>1 => "equilibrium favors Products"

K_C<1 => "equilibrium favors the Reactants"

underline"DATA"

[ICl]_0 = "0.682 g" /"162.357 (g/mol)""= 0.00420 mol"

"Molarity"-> [ICl]_0 = "0.0042 mol" /"0.765 (L)"= 0.00549 M

[I_2]_E = "0.0383 g" /"253.808(g/mol)""= 0.000151 mol"

"Molarity"-> [I_2]_E = "0.000151 mol" /"0.765 (L)"= 0.000197 M = [Cl_2]_E

underline"CONC. AT EQUILIBRIUM"

[ICl]_E = [ICl]_0 - [I_2]_E. - [Cl_2]_E = 0.00549 - 2(0.000197) =

0.005096 M

SUBSTITUTING DATA IN K_C

K_C = ([0.000197] [0.000197])/([0.005096]^2)= color(blue)0.00149