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Bingel [31]
2 years ago
10

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

Chemistry
1 answer:
matrenka [14]2 years ago
5 0

Answer: The value of equilibrium constant for new reaction is 1.92\times 10^{-25}

Explanation:

The given chemical equation follows:

O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)  

The equilibrium constant for the above equation is 5.77\times 10^{-9}

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

3O_2(g)\rightarrow 2O_3(g)

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}

Hence, the value of equilibrium constant for new reaction is 1.92\times 10^{-25}

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if you need to learn more about chemical reaction of Hydrogen sulfide gas, click here

brainly.com/question/15842521?referrer=searchResults

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