Matt214
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Practice 14.2: Predict the equilibrium constant for the first reaction given the equilibrium constants for the second and third reactions: (ans: 1.4 x 102)
CO2 (g) + 3 H2 (g)-><-CH3OH (g) + H2O (g) K1 = ?
CO (g) + H2O (g)-><- CO2 (g) + H2 (g) K2 = 1.0 x 105
CO (g) + 2 H2 (g)-><-CH3OH (g) K3 = 1.4 x 107

Respuesta :

Answer:

K1 = 1.4*10^2

Explanation:

The given reactions are:

Reaction 1

[tex]CO (g) + H2O (g)\rightleftharpoons  CO2 (g) + H2 (g).....K2 = 1.0 * 10^{5}[/tex]

Reaction 2

[tex]CO (g) + 2 H2 (g)\rightleftharpoons CH3OH (g)....K3 = 1.4 * 10^{7}[/tex]

The required reaction is:

[tex]CO2 (g) + 3 H2 (g)\rightleftharpoons CH3OH (g) + H2O (g)... K1 = ?[/tex]

This reaction can be obtained by reversing the first reaction and then adding the reactions 1 and 2.

As per convention

1) For a multistep reaction, the net equilibrium constant is the product of Keq of the individual steps

2) if a reaction is reversed the new equilibrium constant will be the inverse of the old equilibrium constant.

Therefore,

[tex]K1 = \frac{1}{K2}*K3= \frac{1}{1.0*10^{5}}*1.4*10^{7}=1.4*10^{2}[/tex]

Martebi

aThere are different kinds of reaction. The equilibrium constant for the first reaction given the equilibrium constants for the second and third reactions is K1 = 1.4*10^2.

What is equilibrium constant?

This equilibrium constant of a chemical reaction is known to be the sum of the value of a reaction quotient when it is at chemical equilibrium.

It is referred to as a state approached through a dynamic chemical system after only a good and considerate amount of time has elapsed where the composition is said to be no longer of measurable tendency towards any other kind of change.

Learn more about equilibrium constant from

https://brainly.com/question/12858312