Question

Calculate the amounts of each substance in the reaction below if an initial amount of 0.400 moles are brought together with an initial amount of 2.29 moles of Cl2 in a 1.00 L vessel

Answer 1

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

`[COCl_2 ] = 0.282 \ M`

`[CO] = 0.12 \ M`

`[Cl_2] = 2.008 \ M`

Step-by-step explanation:

From the question we are told that

The initial amount of CO is
`n_1 = 0.400`

The initial amount of
`Cl_2` is
`n_2 = 2.20 \ moles`

The volume of the vessel is
`V = 1.0 \ L`

The temperature is
`T = 900 \ K`

The equilibrium constant at the given temperature is
`K_c = 0.800`

The reaction is

`COCl_2 \underset{}{\stackrel{}{\rightleftharpoons}} CO + Cl_2`

Now Generating an I C E table

`COCl_2 \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ \ \ \ \ CO \ \ \ \ \ \ \ \ \ \ + \ \ \ \ Cl_2`

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
`COCl_2` and
`CO` and
`Cl_2` decreased during the reaction

Generally the equilibrium constant is mathematically represented as

`K_c = ([CO] [Cl_2])/([COCl_2])`

=>
`0.800 = ([0.400 - x] [2.20 - x ])/([x])`

=>
`0.800 x = [0.400 - x ] * [ 2.20 - x ]`

=>
`x = 0.282 \ M`

Generally at equilibrium the amount of
`COCl_2` present is

`[COCl_2 ] = 0.282 \ M`

Generally the equilibrium the amount of
`CO` present is

`[CO] = 0.400 - 0.282`

=>
`[CO] = 0.12 \ M`

Generally the equilibrium the amount of
`Cl_2` present is

`[Cl_2 ] = 2.29 - 0.282`

=>
`[Cl_2] = 2.008 \ M`