Question

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [CO2] = 0.24 M, [H2] = 0.24 M, [H2O] = 0.48 M, and [CO] = 0.48 M. Then an additional 0.34 moles per liter of CO2 and H2 are added. When the reaction comes to equilibrium again at the same temperature, what will be the molar concentration of CO?

Answer 1

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

Answer 2

The molar concentration of CO is 1.16 M.

What is molar concentration?

Molar concentration is a measurement of a chemical species concentration in a solution in terms of the amount of substance per unit volume of solution.

The reaction is

$\rm CO_2(g) + H_2(g) <-> H_2O(g) + CO(g)$

Equilibrium concentration of reactant and product is

[CO2] = 0.24 M, [H₂] = 0.24 M, [H2O] = 0.48 M, and [CO] = 0.48 M

$K =\dfrac{[H_2O] [CO] }{[CO_2] [H_2] } \\\\\\K =\dfrac{[ 0.48] [0.48] }{[0.24] [0.24] } =4$

Adding 0.34 to [CO2] = 0.24 M, [H₂] = 0.24 M

(0.24+0.34) M    (0.24+0.34) M

(0.48+x)M         (0.48+x)M

Now the value of K

$4 =\dfrac{ (0.48+x)M (0.48+x)M }{(0.24+0.34) M (0.24+0.34) M } \\x:x = 0.68$

The molar concentration of CO is

[CO] = (0.48+x)M = (0.48+0.68 )M = 1.16 M

Thus, the molar concentration is 1.16 M.

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