Question

consider the following reaction.
hcl(g) + nh₃(g) ⇌ nh₄cl(g)
part 1 of 2
write the expression for the equilibrium constant for this reaction.
k =
part 2 of 2
calculate k using the following concentrations of each substance at equilibrium:
hcl = 0.67 m, nh₃ = 0.51 m, nh₄cl = 2.0 m
round your answer to 2 significant figures.
k =

Explanation:

Part 1 of 2: Write the expression for the equilibrium constant

Step1: Recall the formula for equilibrium constant

For a reaction \(aA + bB
ightleftharpoons cC\), \(K=\frac{[C]^c}{[A]^a[B]^b}\)

Step2: Apply to the given reaction

For \(HCl(g)+NH_3(g)
ightleftharpoons NH_4Cl(g)\), \(a = 1\), \(b = 1\), \(c = 1\). So \(K=\frac{[NH_4Cl]}{[HCl][NH_3]}\)

Part 2 of 2: Calculate \(K\)

Step1: Substitute the values into the formula

Given \([HCl]=0.67\space M\), \([NH_3]=0.51\space M\), \([NH_4Cl]=2.0\space M\)
\(K=\frac{2.0}{0.67\times0.51}\)

Step2: Perform the calculation

\(0.67\times0.51 = 0.3417\)
\(K=\frac{2.0}{0.3417}\approx5.85\)

Step3: Round to 2 significant figures

\(K\approx5.9\)

Answer:

Part 1 of 2: \(K=\frac{[NH_4Cl]}{[HCl][NH_3]}\)
Part 2 of 2: \(K = 5.9\)