Question

Cuso4.5h2o=cuso4.3h2o+2h2o write kp and kc expressions

Answer 1

The Equilibrium Constant For The Reaction As Written Is K =([CuSO4][H2O]5) / [ CuSO4 . 5H2O] K = [H2O]5 K = 5[H2O] K = [H2O]-5 K = [CuSO4 . 5H2O ...

Answer 2

The expressions of $K_p\text{ and }K_c$ are written as.

Explanation:

• Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{p_{C}^cp_{D}^d}{p_{A}^ap_{B}^b}$

Partial pressure of solids and liquids are taken as 1 in the expression of equilibrium constant.

For the given chemical equation:

$CuSO_4.5H_2O(s)\rightleftharpoons CuSO_4.3H_2O(s)+2H_2O(g)$

The expression of $K_p$ for above equation follows:

$K_p=(p_{H_2O})^2$

• Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Concentration of solids and liquids are taken as 1 in the expression of equilibrium constant.

For the given chemical reaction:

$CuSO_4.5H_2O(s)\rightleftharpoons CuSO_4.3H_2O(s)+2H_2O(g)$

The expression of $K_c$ for above equation follows:

$K_c=([H_2O)]^2$

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