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Reil [10]
2 years ago
15

The equilibrium constant for A + 2B → 3C is 2.1 * 10^-6

Chemistry
1 answer:
Mariana [72]2 years ago
7 0

Answer:

b- 4.4 * 10^-12.

Explanation:

Hello.

In this case, as the reaction:

A + 2B → 3C

Has an equilibrium expression of:

K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}

If we analyze the reaction:

2A + 4B → 6C

Which is twice the initial one, the equilibrium expression is:

K_2=\frac{[C]^6}{[A]^2[B]^4}

It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:

K_2=K_1^2

Thus, the equilibrium constant of the second reaction turns out:

K_2=(2.1 * 10^{-6})^2\\\\K_2=4.4x10^{-12}

Therefore, the answer is b- 4.4 * 10^-12.

Best regards.

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