1.25 moles of NOCl were placed in a 2.50 L reaction chamber at 427ºC. Afterequilibrium was reached, 1.10 moles of NOCl remained. Calculate the equilibriumconstant, Kc, for the reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g)

Answer 1

The balanced equation for the reaction is:

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

Let x be the amount (in moles) of NO and Cl2 that are formed at equilibrium.

Then, the amount of NOCl remaining at equilibrium is (1.25 - x) moles.

The equilibrium expression for the reaction is:

Kc = [NO]^2[Cl2] / [NOCl]^2

At equilibrium, the concentration of NO and Cl2 is each x/2 moles per liter, and the concentration of NOCl is (1.25 - x) / 2.5 moles per liter.

Substituting these values into the equilibrium expression gives:

Kc = [(x/2)^2 * (x/2)] / [((1.25 - x)/2.5)^2]

Simplifying this expression gives:

Kc = (x^3) / (5.625 - 5x + x^2)

Now we can solve for x using the given information that 1.10 moles of NOCl remain at equilibrium:

1.10 = 1.25 - x
x = 0.15 moles

Substituting this value of x into the expression for Kc gives:

Kc = (0.15^3) / (5.625 - 0.75 + 0.0225)
Kc = 0.0062

Therefore, the equilibrium constant, Kc, for the reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g) at 427ºC is 0.0062.