Question

The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K.

NH4Cl(s) <---> NH3(g) + HCl(g)

Calculate the equilibrium concentration of HCl when 0.338 moles of NH4Cl(s) are introduced into a 1.00 L vessel at 548 K.

[arrows used to indicate rxn]

Answer 1

ANSWER:

Given,

K_c = 5.10×10^-6

number of moles of NH₄Cl = 0.338 moles

volume of solution = 1.00 L

concentration of NH₄Cl ={number of moles of NH₄Cl}/{volume of solution in L}

= {0.338 mol}/{1.00 L}

= 0.338 M

Now,

For, equilibrium constant:

,

We will take x = 1.31 x 10^-3 for this calculation because concentration can not be negative.

then,

equilibrium concentration of HCl = 1.31 x 10^-3​ M

equilibrium concentration of NH₃ = 1.31 x 10^-3​ M

equilibrium concentration of NH₄Cl = 0.3367​ M