Question

at a particular temperature, a 3.0 - l flask contains 1.5 moles of cl₂, 1.6 moles of nocl, and 4.5×10⁻³ mole of no. calculate k at this temperature for the following reaction. 2nocl(g) ↔ 2no(g) + cl₂(g)

Explanation:

Step1: Calculate concentrations

Concentration \(c=\frac{n}{V}\).
For \(Cl_2\): \(c_{Cl_2}=\frac{1.5\space mol}{3.0\space L}= 0.50\space M\)
For \(NOCl\): \(c_{NOCl}=\frac{1.6\space mol}{3.0\space L}\approx0.53\space M\)
For \(NO\): \(c_{NO}=\frac{4.5\times 10^{-3}\space mol}{3.0\space L}=1.5\times 10^{-3}\space M\)

Step2: Write the equilibrium constant expression

For the reaction \(2NOCl(g)
ightleftharpoons 2NO(g)+Cl_2(g)\), the equilibrium constant \(K = \frac{[NO]^2[Cl_2]}{[NOCl]^2}\)

Step3: Substitute concentrations into the expression

\(K=\frac{(1.5\times 10^{-3})^2\times0.50}{(0.53)^2}\)
First, calculate \((1.5\times 10^{-3})^2 = 2.25\times 10^{-6}\)
Then, \(2.25\times 10^{-6}\times0.50=1.125\times 10^{-6}\)
And \((0.53)^2 = 0.2809\)
\(K=\frac{1.125\times 10^{-6}}{0.2809}\approx4.0\times 10^{-6}\)

Answer:

\(4.0\times 10^{-6}\)