Chemistry College

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?

Answers

Answer 1

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

Answer 2

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]  

What is equilibrium constant?

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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Related Questions

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