Question

For this heterogeneous system 2A(aq)+3B(g)+C(l) ↽− −⇀ 2D(s)+3E(g) 2A(aq)+3B(g)+C(l)↽−−⇀2D(s)+3E(g) the concentrations and pressures at equilibrium are [A]=2.71× 10 −2 M [A]=2.71×10−2 M , P B =7.00× 10 3 Pa PB=7.00×103 Pa , [C]=11.54 M [C]=11.54 M , [D]=13.14 M [D]=13.14 M , and P E =5.03× 10 4 torr PE=5.03×104 torr . Calculate the thermodynamic equilibrium constant, K K .

Answer 1

Answer : The value of thermodynamic equilibrium constant (K) is, $7.56\times 10^6$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$2A(aq)+3B(g)+C(l)\rightleftharpoons 2D(s)+3E(g)$

The expression of $K$ will be,

$K=\frac{[D]^2[E]^3}{[A]^2[B]^3[C]}$

Given:

[A] = $2.71\times 10^{-2}M$

[B] = $7.00\times 10^{3}M$

[C] = $11.54M$

[D] = $13.14M$

[E] = $5.03\times 10^{4}M$

Now put all the given values in the above formula expression, we get:

$K=\frac{(13.14)^2\times (5.03\times 10^{4})^3}{(2.71\times 10^{-2})^2\times (7.00\times 10^{3})^3\times (11.54)}$

$K=7.56\times 10^6$

Therefore, the value of thermodynamic equilibrium constant (K) is, $7.56\times 10^6$