Question

for the reaction given below, 2.00 moles of a and 3.00 moles of b are placed in a 6 l container.
a(g) + 2b(g) ⇋ c(g)
at equilibrium, the concentration of a is 0.238 mol/l.
what is the value of k, reported to 3 significant figures?

Explanation:

Step1: Calculate initial concentrations

Initial concentration of \(A\), \([A]_0=\frac{2.00\ mol}{6\ L}=\frac{1}{3}\ mol/L\approx0.333\ mol/L\)
Initial concentration of \(B\), \([B]_0=\frac{3.00\ mol}{6\ L}= 0.5\ mol/L\)

Step2: Find the change in concentration of \(A\)

Let the change in concentration of \(A\) be \(x\). At equilibrium, \([A]=[A]_0 - x\). Given \([A]=0.238\ mol/L\)
\(x=[A]_0 - [A]=0.333 - 0.238 = 0.095\ mol/L\)

Step3: Calculate equilibrium concentrations of \(B\) and \(C\)

From the stoichiometry of the reaction \(A(g)+2B(g)
ightleftharpoons C(g)\), change in concentration of \(B\) is \(2x\) and change in concentration of \(C\) is \(x\)
Equilibrium concentration of \(B\), \([B]=[B]_0-2x=0.5 - 2\times0.095=0.5 - 0.19 = 0.31\ mol/L\)
Equilibrium concentration of \(C\), \([C]=x = 0.095\ mol/L\)

Step4: Calculate the equilibrium constant \(K\)

The equilibrium constant expression for the reaction \(K=\frac{[C]}{[A][B]^2}\)
Substitute the values: \(K=\frac{0.095}{0.238\times(0.31)^2}\)
First calculate \((0.31)^2 = 0.0961\)
Then \(0.238\times0.0961=0.0228718\)
\(K=\frac{0.095}{0.0228718}\approx4.15\)

Answer:

\(4.15\)