Question

The equilibrium constant, K, for the following reaction is 2.57×10-4 at 549 K.

NH4Cl(s) NH3(g) + HCl(g)

An equilibrium mixture in a 14.3 L container at 549 K contains 0.321 mol NH4Cl(s), 1.97×10-2 M NH3 and 1.30×10-2 M HCl.

What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 6.96 L?

[NH3] = __________M

[HCl] = ___________M

Answer 1

Answer:

NH4Cl(sNH3(g) + HCl(g)

Initially,molar grouping of NH3 = moles of NH3/Volume of soltion in liters = 0.0197/6.96 = 0.00273 M

Initially,molar grouping of HCl = moles of HCl/Volume of soltion in liters = 0.0130/6.96 = 0.00187 M

Therefore, Reaction Quotient,Kq = [NH3]*[HCl] = 5.1051*10-6

Since, Kq < Keqb , therefore, more items will be shaped

Let at eqb., [NH3] = 0.00273+x and [HCl] = 0.00187+x

In this way, Keq = [NH3]*[HCl] = (0.00273+x)*(0.00187+x)

or on the other hand, 2.57*10^-4 = x2 + 0.0046x+0.0000051051

Subsequently, x2 + 0.0046x - 0.000252 = 0

or then again, x = 0.01374

Subsequently, at eqb., [NH3] = 0.01647 M

[HCl] = 0.01561 M