Part II. Long questions (38 pt]: Clearly show explanation 1. For the equilibrium

Question

Part II. Long questions (38 pt]: Clearly show explanation 1. For the equilibrium NHCI (s)NH, (g)+HCI(e). K- 25 at 29 your detailed explanation and calculation to receive partial credits. (a) (6pt) If we mix 2.0-m and allowed to come to equilibrium, determine t ol NH, (g) and 2.0-mol HCI (s) in a 1.00L container (without NH CI (s) in the he equilibrium concentration of NHs (8) and Hci (8) (s) in the beginning) at 293C b) (4pt) Now let's onsider a different case but with the same Kc. At equilbrium, we have (B) and HCl (g) at 293 °C. If the concentration of NH, () is 0.52 mol/ and the an concentration of HCI (8)? c i10, what is the .20 g, what is the ount of NHCi (s) is 1 (c) (3pt) Determine the equilibrium constant of the following reaction 4 NH, (0) +4 Hc (

Explanation / Answer

1. For the reaction,

NH4Cl(s) <==> NH3(g) + HCl(g)   Kc = 25

(a) Initial [NH3(g)] = 2.0 mol/1 L = 2 M

Initial [HCl(g)] = 2.0 mol/1 L = 2.0 M

ICE chart

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

Initial                           -                    2.0           2.0

Change                     +x                    -x             -x

Equilibrium                  x                 2.0 - x     2.0 - x    

So,

Kc = [NH3][HCl]/[NH4Cl]

25 = x/(2-x)(2-x)

25 = x/(4 - 4x + x^2)

24x^2 - 100x + 100 = 0

x = 1.7 M

So,

equilibrium [NH3(g)] = 1.7 M

equilibrium [HCl(g)] = 1.7 M

(b) at equilibrium [NH3(g)] = 0.52 M

at equilibrium = [NH4Cl] = 1.20 g/53.5 g/mol x 1 L = 0.022 M

So,

Kc = [NH3][HCl]/[NH4Cl]

25 = 0.022/0.52[HCl]

at equilibrium [HCl(g)] = 0.0017 M

(c) For the reaction,

4NH3(g) + 4HCl(g) <==> 4NH4Cl

So equilibrium constant Kc,

Kc = 1/25 = 0.04

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