Question

Suppose that for a hypothetical reaction: A2(g) + 3B2(g) 2AB3(g) It is determined that at a certain temperature the equilibrium concentrations are [A2] = 0.0055, [B2] = 0.763 and [AB3] = 7.6×10-4. Calculate the numerical value of K for the reaction.

Answer 1

Answer:

The numerical value of K for the reaction is 2.36*10⁻⁴

Explanation:

A chemical equilibrium is a reaction that is never completed, since it occurs simultaneously in both directions (the reactants form products, and in turn, these form reactants again). In other words, it is a dynamic balance. When the concentrations of each of the substances involved (reactants or products) stabilize, that is, they are spent at the same rate as they are formed, chemical equilibrium is reached.

The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations of reactants and products. Its value in a chemical reaction depends on the temperature, so this must always be specified.

Being:

aA + bB ⇄ cC + dD

then the constant Kc is:

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

In the case of the reaction:

A₂(g) + 3 B₂(g) ⇒ 2 AB₃(g)

The constant Kc is:

$Kc=\frac{[AB_{3} ]^{2} }{[A_{2} ] *[B_{2} ]^{3} }$

Being:

• [AB₃]=7.6*10⁻⁴
• [A₂]= 0.0055
• [B₂]= 0.763

and replacing, you get:

$Kc=\frac{(7.6*10^{-4} )^{2} }{0.0055 *(0.763)^{3} }$

Kc=2.36*10⁻⁴

The numerical value of K for the reaction is 2.36*10⁻⁴