Question

1. if initially there is 38.0 mol/l of so₃(g), at equilibrium there is 14.0 mol/l of so₂(g). calculate kc.

2so₂(g) + o₂(g) ⇌ 2so₃(g)

Explanation:

Step1: Determine the change in concentration of \(SO_3\)

The initial concentration of \(SO_3\) is \(38.0\ mol/L\) and at equilibrium the concentration of \(SO_2\) is \(14.0\ mol/L\). From the balanced - equation \(2SO_2(g)+O_2(g)
ightleftharpoons 2SO_3(g)\), for every 2 moles of \(SO_2\) produced, 2 moles of \(SO_3\) are consumed. The change in concentration of \(SO_3\) (\(\Delta[SO_3]\)) is equal to the concentration of \(SO_2\) produced since the stoichiometric coefficients of \(SO_2\) and \(SO_3\) are the same. So, \(\Delta[SO_3]=14.0\ mol/L\). The equilibrium concentration of \(SO_3\) is \([SO_3]=38.0 - 14.0=24.0\ mol/L\).

Step2: Determine the equilibrium concentration of \(O_2\)

From the balanced - equation, the stoichiometric ratio of \(SO_2\) to \(O_2\) is \(2:1\). So, if \([SO_2] = 14.0\ mol/L\) at equilibrium, then \([O_2]=\frac{14.0}{2}=7.0\ mol/L\) at equilibrium.

Step3: Write the expression for \(K_c\)

The equilibrium constant expression for the reaction \(2SO_2(g)+O_2(g)
ightleftharpoons 2SO_3(g)\) is \(K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}\).

Step4: Substitute the equilibrium concentrations into the \(K_c\) expression

Substitute \([SO_3]=24.0\ mol/L\), \([SO_2]=14.0\ mol/L\), and \([O_2]=7.0\ mol/L\) into the \(K_c\) expression:
\[K_c=\frac{(24.0)^2}{(14.0)^2\times7.0}=\frac{576}{196\times7.0}=\frac{576}{1372}\approx0.42\]

Answer:

\(K_c\approx0.42\)