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
Jose
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#

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