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kow [346]
4 years ago
11

Write the expression for the equilibrium constant for the reaction represented by the equation CaCO3(s)⇌Ca2+(aq)+CO32−(aq). Is K

c > 1, < 1, or ≈ 1? Explain your answer.
Chemistry
1 answer:
Alexeev081 [22]4 years ago
5 0

<u>Answer:</u> For the given reaction, the value of K_c is greater than 1

<u>Explanation:</u>

For the given chemical equation:

CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)

The expression of K_c for above equation follows:

K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}]

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression

As, the denominator is missing and the numerator is the only part left in the expression. So, the value of K_c will be greater than 1.

Hence, for the given reaction, the value of K_c is greater than 1

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