Complete Question
Calculate the equilibrium amounts of each substance in the reaction below if an initial amount of 0.400 moles of CO are brought together with an initial amount of 2.20 moles of Cl2 in a 1.00 L vessel and then equilibrium is established at 900 K. Kc at this temperature = 0.800. COCl2(g) CO(g) + Cl2(g)
Answer:
The amount of each substance at equilibrium is
[tex][COCl_2 ] = 0.282 \ M[/tex]
[tex][CO] = 0.12 \ M[/tex]
[tex][Cl_2] = 2.008 \ M[/tex]
Explanation:
From the question we are told that
The initial amount of CO is [tex]n_1 = 0.400[/tex]
The initial amount of [tex]Cl_2[/tex] is [tex]n_2 = 2.20 \ moles[/tex]
The volume of the vessel is [tex]V = 1.0 \ L[/tex]
The temperature is [tex]T = 900 \ K[/tex]
The equilibrium constant at the given temperature is [tex]K_c = 0.800[/tex]
The reaction is
[tex]COCl_2 \underset{}{\stackrel{}{\rightleftharpoons}} CO + Cl_2[/tex]
Now Generating an I C E table
[tex]COCl_2 \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ \ \ \ \ CO \ \ \ \ \ \ \ \ \ \ + \ \ \ \ Cl_2[/tex]
Initial [I] 0 0.400 2.20
Change [C ] + x -x - x
Equilibrium [E ] x 0.400- x 2.20 - x
Here x is the amount in terms of concentration by which [tex]COCl_2[/tex] and [tex]CO[/tex] and [tex]Cl_2[/tex] decreased during the reaction
Generally the equilibrium constant is mathematically represented as
[tex]K_c = \frac{[CO] [Cl_2]}{[COCl_2]}[/tex]
=> [tex]0.800 = \frac{[0.400 - x] [2.20 - x ]}{[x]}[/tex]
=> [tex]0.800 x = [0.400 - x ] * [ 2.20 - x ][/tex]
=> [tex]x = 0.282 \ M[/tex]
Generally at equilibrium the amount of [tex]COCl_2[/tex] present is
[tex][COCl_2 ] = 0.282 \ M[/tex]
Generally the equilibrium the amount of [tex]CO[/tex] present is
[tex][CO] = 0.400 - 0.282[/tex]
=> [tex][CO] = 0.12 \ M[/tex]
Generally the equilibrium the amount of [tex]Cl_2[/tex] present is
[tex][Cl_2 ] = 2.29 - 0.282[/tex]
=> [tex][Cl_2] = 2.008 \ M[/tex]
