A sealed reaction vessel initially contains 1.113×10-2 moles of water vapor and 1.490×10-2 moles of CO(g).

H20(g) +CO(g) <==> H2(g) + CO2(g)

After the above reaction has come to equilibrium, the vessel contains 8.326×10-3 moles of CO2(g).

What is the value of the equilibrium constant Kc of the reaction at the temperature of the vessel?

Answer:

Kc =[tex]\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }[/tex]

Kc = 50.2059

Explanation:

1. Balance the equation

2. Use the Kc formula

Remember that pure substances, like H2 are not included on the Kc formula

The value of the equilibrium constant Kc of the reaction at the temperature of the vessel is 0.05.

Calculation,

The balanced chemical equation is given as,

                      [tex]2H_{2}O[/tex]             +         [tex]2CO[/tex]            →            [tex]2H_{2}[/tex]       +     [tex]2CO_{2}[/tex]

At t=0          1.113×[tex]10^{-2}[/tex]                 1.490×[tex]10^{-2}[/tex]                                         0

At [tex]t=t_{eq}[/tex] 1.113×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex]   1.490×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex]            8.326×[tex]10^{-3}[/tex]  

The number of moles of [tex]H_{2}O[/tex] at equilibrium  =  1.113×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex]

The number of moles of [tex]CO[/tex] at equilibrium  = 1.490×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex]

The number of moles of  [tex]CO_{2}[/tex]   at equilibrium =   8.326×[tex]10^{-3}[/tex]  

Equilibrium constant is calculated bt ratio of product of concentration of reactant by product.

K = [tex][H_{2}O]^{2}[/tex]×[tex][CO]^{2}[/tex]/[tex][CO_{2}]^{2}[/tex]

K =( 1.113×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex] )∧2× (1.490×[tex]10^{-2}[/tex]-8.326×[tex]10^{-3}[/tex])∧2/(8.326×[tex]10^{-3}[/tex])∧2

K = 0.049 ≈ 0.05

To learn more about equilibrium constant,

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