Question

1.25 mol NOCl was placed in a 2.50 L reaction chamber at 427°C. After equilibrium was reached, 1.10 moles of NOCl remained. Calculate the equilibrium constant Kc for the reaction.

2NOCl(g)= 2NO(g) + Cl2(g) 

Answer 1

Q161077

1.25 mol NOCl was placed in a 2.50 L reaction chamber at 427°C. After equilibrium was reached, 1.10 moles of NOCl remained. Calculate the equilibrium constant Kc for the reaction.

2NOCl(g)= 2NO(g) + Cl₂(g) 

Solution:

1.25 mol NOCl was placed in a 2.50 L reaction chamber at 427°C.

So the initial concentration of NOCl is

[NOCl] _initial $= \frac{initial \space moles \space of \space NOCl}{volume\space of \space container\space in\space 'L' }$ $= \frac{1.25 \space mol \space NOCl}{2.50 \space L}$

$= 0.5 \space mol/L$

When the reaction reached equilibrium, 1.10 moles of NOCl remained in the chamber, so the

concentration of NOCl at equilibrium is

[NOCl]equilibrium $= \frac{equilibrium \space moles \space of \space NOCl}{volume\space of \space container\space in\space 'L' }$ $= \frac{1.10 \space mol \space NOCl}{2.50 \space L}$

$= 0.44 \space mol/L$

Next we will draw the ICE table for the given reaction.

I will show the image of the ICE table I have drawn.

So we have

[NOCl] _equilibrium = 0.5 – x = 0.5 – 2 * 0.03  = 0.5 – 0.06 = 0.44 mol/L

[NO] _equilibrium  =  2x = 2 * 0.03   = 0.06 mol/L

[Cl2] _equilibrium = x = 0.03 mol/L 

Now consider the reaction

2NOCl (g) <==> 2NO(g) + Cl₂(g)

equilibrium constant equation for this reaction is written as

$Kc = \frac{[NO]^2*[[Cl_2]}{[NOCl]^2}$ ;

plug the equilibrium concentration in this equation we have

$Kc = \frac{[0.06mol/L]^2*0.03mol/L}{[0.44mol/L]^2}$ $= \frac{0.0036*0.03 }{0.1936} = \frac{0.000108}{0.1936} = 0.0005579 mol/L$

which can also be written as Kc = 5.58 * 10^-4 mol/L ;

So the equilibrium constant, Kc for the given reaction is 5.58 * 10^-4 mol/L