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H2(g) + CO2(g) + 10 kcal H2O(g) + CO(g)

In another experiment involving the above reaction at 1,000°C, the equilibrium concentrations were found to be:

H2(g) = 2.0 moles per liter
H2O(g) = 4.0 moles per liter
CO2(g) = 5.0 moles per liter
CO(g) = 4.0 moles per liter

What is the equilibrium constant, Keq, for this reaction?

Respuesta :

Eshaunlee476
This might be a little hard for me but I got it. So you have 2.0, 4.0, 5.0 and 4.0. Subtract from 3.0 - 2.0 = 1.0 and 4.5 - 2.0 = 2.5.

So now we have other answers. 

H20, H2, CO2 and CO

Keq reaction for each formula is 1.6

Welcome :)

Answer: The equilibrium constant for the given reaction is 0.8

Explanation:

Equilibrium constant is defined as the ratio of concentration of the products raised to the power its stoichiometric coefficients to the concentration of reactants raised to power its stoichiometric coefficient. It is represented as [tex]K_c[/tex]

For the general equation:

[tex]aA+bB\rightleftharpoons cC+dD[/tex]

The equilibrium constant is represented as:

[tex]K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

For the given chemical equation:

[tex]H_2(g)+CO_2(g)+10 kcal\rightleftharpoons H_2O(g)+CO(g)[/tex]

[tex]K_c[/tex] for this equation is given by:

[tex]K_c=\frac{[H_2O][CO]}{[H_2][CO_2]}[/tex]

Concentration at equilibrium of

[tex]H_2=2mol/L\\H_2O=4mol/L\\CO_2=5mol/L\\CO=4mol/L[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{4\times 4}{2\times 5}\\K_c=0.8[/tex]

Hence, the equilibrium constant for the given chemical reaction is 0.8