Question

Пожалуйста помогите решить!!написать выражение константы равновесия для3А+B<->A3B

Answer 1

$K=\frac{[A_{3}B]}{[A]^{3}*[B]}$

Answer 2

(A3B)^2/ (3*A)*B из формулы Константы равновесия : Кр= P^2вещества/ Р(первого реагента) * Р(второго реагента)

Answer 3

Writing the Expression for the Equilibrium Constant

To write the expression for the equilibrium constant, we can use the following general equation for a reversible reaction:

aA + bB ⇌ cC + dD

In this case, we have the reaction:

3A + B ⇌ A3B

The equilibrium constant expression (Kc) for this reaction can be written using the concentrations of the reactants and products. The general form of the equilibrium constant expression is:

Kc = [C]^c[D]^d / [A]^a[B]^b

Where: - [C], [D], [A], and [B] are the molar concentrations of the products and reactants at equilibrium. - a, b, c, and d are the coefficients of the balanced chemical equation.

Writing the Equilibrium Constant Expression for 3A + B ⇌ A3B

For the given reaction 3A + B ⇌ A3B, the equilibrium constant expression (Kc) can be written as follows:

Kc = [A3B] / [A]^3[B]

Where: - [A3B] represents the molar concentration of A3B at equilibrium. - [A] represents the molar concentration of A at equilibrium. - [B] represents the molar concentration of B at equilibrium.

This expression represents the equilibrium constant for the given reaction 3A + B ⇌ A3B.