Question

Compounds A, B, and C react according to the following equation. 3A(g) + 2B(g) 2C(g) At 100°C a mixture of these gases at equilibrium showed that [A] = 0.855 M, [B] = 1.23 M, and [C] = 1.75 M. What is the value of Kc for this reaction?

Answer 1

Answer:

The value of Kc for the reaction is 3.24

Explanation:

A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products ( direct reaction) and products transforming back into reagents (inverse reaction)

Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.

Being:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:

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

Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.

Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g),  the constant Kc is:

$Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }$

where:

• [A]= 0.855 M
• [B]= 1.23 M
• [C]= 1.75 M

Replacing:

$Kc=\frac{1.75^{2} }{0.855^{3}*1.23^{2} }$

Solving you get:

Kc=3.24

The value of Kc for the reaction is 3.24