Question

\(2NOCl(g) ⇌ 2NO(g) + Cl_2(g)\)
In an experiment, \(2.0\) moles of \(NOCl\) was placed in a one-litre flask and the concentration of \(NO\) after equilibrium established, was found to be \(0.4\) mol/ L. The equilibrium constant at \(30°C\) is ____ \(×10^{–4}\).

Answer 1

Correct Answer: 125

 To calculate the equilibrium constant \(K_c\) for the reaction \(2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)\), we begin by setting up an ICE (Initial, Change, Equilibrium) table.

Species	Initial (mol/L)	Change (mol/L)	Equilibrium (mol/L)
\(NOCl\)	2.0	-2x	2.0-2x
\(NO\)	0	+2x	2x
\(Cl_2\)	0	+x	x

Given: \(2x = 0.4\) mol/L, thus \(x = 0.2\) mol/L. Now use these values to find the equilibrium concentrations:

• \([NOCl] = 2.0 - 2(0.2) = 1.6\) mol/L
• \([NO] = 0.4\) mol/L
• \([Cl_2] = 0.2\) mol/L

Substitute these into the expression for \(K_c\):

\(K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = \frac{(0.4)^2(0.2)}{(1.6)^2}\)

\(= \frac{0.16 \cdot 0.2}{2.56}\)

\(= \frac{0.032}{2.56} = 0.0125\)

Thus, \(K_c = 0.0125 \times 10^{0} = 125 \times 10^{-4}\), which fits the provided range of 125.