WS 7 FR Equil Kc Kp Ksp Q KEY PDF

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WS 7 FR Equil Kc Kp Ksp Q KEY...

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WS FR Equilibrium (Kc, Kp, Ksp, Q) CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS. It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do not. Attention should be paid to significant figures. 1.

2 H2S(g) ↔ 2 H2(g) + S2(g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72 x 10–2 mol of S2(g) is present at equilibrium. (a)

Write the equilibrium expression constant, Kc, for the decomposition reaction represented above. (1) Kc = [H2]2[S2] [H2S]2

(b)

Calculate the equilibrium concentration, in mol∙L–1 (M), of the following gases in the container at 483 K. (i)

H2(g)

(1)

(ii)

H2S(g)

(2)

(initial)

[H2S]initial = 3.40 g H2S x 1 mol H2S = 0.0998 mol H 2S

0.0998 mol H2S = 0.0798 M H2S

34.08 g H2S

1.25 L

[S2]eq = (0.0372 mol) = 0.0298 M S2 (at equilibrium) 1.25 L

2 H2S(g) I 0.0798 M C –0.0595 E 0.0202 M

↔

2 H2(g) + S2(g) 0M 0M +0.0595 +0.0298 0.0595 M 0.0298 M

[H2]eq = 0.0595 M [H2S]eq = 0.0202 M (as shown is (b)(i) above)

1

(c)

Calculate the value of the equilibrium constant, Kc, for this reaction. (2) Kc = [H2]2[S2] [H2S]2

Kc = (0.0595)2(0.0372 / 1.25) = 0.258 (0.0202)2

(d)

Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K. (2) PS2 = ? nS2 = 3.72 x 10–2 mol V = 1.25 L T = 483 K

PV = nRT P(1.25 L) = (3.72 x 10–2)(0.08206)(483) P = 1.18 atm S2

(e)

For the reaction H2(g) + ½ S2(g) ↔ H2S(g) constant, Kc. (2) Given:

Kc =

at 483 K, calculate the equilibrium

2 H2S(g) ↔ 2 H2(g) + S2(g)

Kc = 0.258

H2(g) + ½ S2(g) ↔ H2S(g)

Kc = ?

1 = 1 √Kc √0.258

= 1.97

2

N2(g) + 3 H2(g) ↔ 2 NH3(g) 2.

For the reaction represented above, the equilibrium constant, Kp, is 3.1 x 10–4 at 700 K. (a)

Write the expression for the equilibrium constant, Kp, for the reaction. (2) Kp =

(b)

(PNH3)2 (PN2)(PH2)3

Assume that the initial partial pressures of the gases are as follows: PN2 = 0.411 atm, PH2 = 0.903 atm, and PNH3 = 0.224 atm. (i)

Calculate the value of the reaction quotient, Q, at these conditions. (1) Q =

(ii)

(PNH3)2 (PN2)(PH2)3

Q =

Q = 0.166 (0.224)2 (0.411)(0.903) 3

The system is allowed to reach equilibrium under these conditions. Determine whether the equilibrium concentration of NH3(g) will be greater than, equal to, or less than the initial concentration of NH3(g) . Justify your answer. (2) Q > Kp , there is too much product initially so the reaction will shift left to reach equilibrium by making more reactant and consuming product. The equilibrium [NH3] will be less than the initial.

(c)

The value of the equilibrium constant, Kp, at 300 K is 4.3 x 10–2. Is the reaction endothermic or exothermic. Justify you answer. (2) As the temperature decreased from 700 K to 300 K, the value of Kp increased. Since Kp increased the reaction must have shifted right to produce more products. Since the reaction shifted right by decreasing temperature (removing heat), it must be exothermic.

(d)

The value of Kp for the reaction represented below is 8.3 x 10–3 at 700 K. NH3(g) + H2S(g) ↔ NH4HS(g) Calculate the value of Kp at 700 K for each of the reactions represented below. (i)

NH4HS(g) ↔ NH3(g) + H2S(g) Kp =

1

(1) = 1.2 × 10

2

8.3 x 10−3

(ii)

2 H2S(g) + N2(g) + 3 H2(g) ↔ 2 NH4HS(g) 2 × [NH3(g) + H2S(g) ↔ NH4HS(g)] N2(g) + 3 H2(g) ↔ 2 NH3(g)

(2)

Kp = (8.3 × 10−3)2 Kp = 3.1 × 10−4

3

Kp = (8.3 × 10−3)2 (3.1 × 10−4) = 2.1 × 10−8

2 H2S(g) + N2(g) + 3 H2(g) ↔ 2 NH4HS(g)

3.

Answer the following questions relating to the solubility of the chlorides of silver and lead. (a) At 10oC, 8.9 x 10–5 g AgCl(s) will dissolve in 100 mL of water. (i)

Write the equation for the dissociation of AgCl(s) in water. (1) Ag+(aq) + Cl(aq)

AgCl(s)

(ii)

Calculate the solubility, in mol L–1, of AgCl(s) in water at 10oC. (2) 8.9 x 10–5 g x 1 mol = 6.2 x 10–7 mol 143.32 g 6.2 10–7 mol = 6.2 x10–6 M AgCl 0.100 L

(iii)

Calculate the value of the solubility product constant, Ksp, of AgCl(s) at 10oC. (1)

I

AgCl(s)

Ag+

6.2 x 10–6 M

0M

C –6.2 x 10–6 M x 10–11

Cl–

+

Ksp = [Ag+][Cl–] Ksp = (6.2 x 10–6)2

0M

+6.2 x 10–6 M +6.2 x 10–6 M

Ksp = 3.8

(b) At 25oC, the value of Ksp for PbCl2(s) is 1.6 x 10–5 and the Ksp for AgCl(s) is 1.8 x 10–10. (i) Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated PbCl2 solution to which 0.250 mole NaCl(s) has been added. Assume no change in volume. (1) PbCl2(s) I

x

C

–x

E

0M

Pb2+

+

0M

2 Cl–

Ksp = 1.6 x 10–5 Ksp = [Pb2+][Cl–]2

0.250 M

+x

+x

x

0.250 + x

1.6 x 10–5 = [Pb2+](0.250)2 [Pb2+] = 2.6 x 10–4 M

(ii) If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M Pb(NO3)2(aq), will a precipitate form? Assume that volumes are additive. Show calculations to support your answer. (3) Pb2+ + 2 Cl–

PbCl2(s)

Ksp = [Pb2+][Cl–]2 Ksp = 1.6 x 10–5

[Pb2+] = _?_ (after mixing):

M 1 V1 = M 2 V2

2+

(0.030 M Pb )(60.0 mL) = MPb2+ (120 mL) [Pb2+] = 0.015 M

Q = [Pb2+][Cl–]2 Q = (0.015)(0.020) 2 = 6.0 x 10–6 Q < K, therefore a precipitate will not 4 form as the reaction proceeds to the right

[Cl–] = _?_ (after mixing):

M 1 V1 = M 2 V2

–

(0.040 M Cl )(60.0 mL) = MCl– (120 mL) [Cl–] = 0.020 M

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