Question

Calculate the equilibrium constant for the reaction below given that [Cl2] = 0.37 M, [H2] = 0.27 M, and [HCl] = 0.95 M at equilibrium. Given: Cl2(g) + H2(g) ⇌ 2HCl(g)

Answer 1

Answer:

The equilibrium constant is 9.034.

Explanation:

Every reversible chemical reaction, as in this case, occurs in both directions: the reagents are transformed into products (direct reaction) and the products are transformed back into reagents ( reverse reaction).

The general way in which a reversible reaction can be written is:

aA + bB ⇔ cC + dD

where A, B, C and D represent the chemical species involved and a, b, c and d their respective stoichiometric coefficients.

The chemical equilibrium is the state in which the direct and indirect reaction have the same reaction rate, and is expressed by a constant Kc. This constant is defined as:

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

This constant is equal to the multiplication of the concentrations of the products elevated to their stoichiometric coefficients divided the multiplication of the concentrations of the reactants elevated to their stoichiometric coefficients. In Kc only gases and aqueous solutions come into play and only depends on the temperature.

You have the reaction:

Cl₂(g) + H₂(g) ⇌ 2 HCl(g)

So, in this case, the constant Kc is:

$Kc=\frac{[HCl]^{2} }{[Cl_{2} ]*[H_{2}] }$

Then:

$Kc=\frac{(0.95 M)^{2} }{0.37 M*0.27 M}$

Kc= 9.034

The equilibrium constant is 9.034.