Download Chapter 13 – Equilibrium Calculations 4

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Chapter 13 – Equilibrium Calculations 4
1. For the following reaction, Kp = 0.262 at 1000ºC: C(s) + 2 H2(g)  CH4(g)
At equilibrium, PH2 is 1.22 atm. What is the equilibrium partial pressure of CH4(g)?
K = 0.262 = PCH4 / PH22 = PCH4 / (1.22)2
PCH4 = 0.390 atm
2. When 2.50 M carbon dioxide gas is sealed in a container, some decomposes to carbon
monoxide and oxygen gases. Kc for this reaction at 250°C is 2.0 x 10–6. Calculate the
equilibrium concentration of each component.
2 CO2(g) 
2 CO(g) +
O2(g)
2.50
–2x
2.50 – 2x
0
+2x
2x
0
+x
x
(2x)2 x / (2.50)2 = 2.0 x 10–6
x = [O2] = 0.015 M
[CO] = 0.030 M
[CO2] = 2.47 M
K is small – use shortcut!
3. Ammonium hydrogen sulfide decomposes according to the following reaction, for which
Kp = 0.11 at 250ºC:
NH4HS(s)  H2S(g) + NH3(g)
If 55.0 g NH4HS(s) is placed in a sealed 5.0 L container, what is the partial pressure of NH3(g) at
equilibrium?
I
C
E
NH4HS(s)  H2S(g)
0
+x
x
x2 = 0.11
+
NH3(g)
0
+x
x
x = PNH3 = 0.33 atm
4. When nitric oxide (NO) decomposes to give oxygen and nitrogen, Kc = 2.3 x 1030 at 298 K.
In the atmosphere, PO2 = 0.209 atm and PN2 = 0.781 atm. What is the equilibrium partial
pressure of NO in the air we breathe?
Kp = Kc (RT)Δn
E
2 NO(g) 
x
N2(g)
0.781
Δn = 0, so Kp = Kc
+
O2(g)
0.209
2.3 x 1030 = (0.781)(0.209) / (PNO2)
PNO = 2.7 x 10–16 atm
5. When chlorine gas is placed in a container with H2O(g), some gaseous HCl and oxygen is
formed. Kp = 3.2 x 10–14 for the reaction at 25°C. Calculate the pressure of each component at
equilibrium if the initial pressures of chlorine and water vapor are 0.80 atm and 0.40 atm,
respectively.
I
C
E
2 Cl2(g) +
0.80
–2x
0.80–2x
2 H2O(g) 
0.40
–2x
0.40–2x
4 HCl(g) +
0
+4x
4x
O2(g)
0
+x
x
K is small – use shortcut!
(4x)4 x / (0.80)2 (0.40)2 = 3.2 x 10–14
x = PO2 = 4.2 x 10–4 atm
PHCl = 1.7 x 10–3 atm
PCl2 = 0.80 atm
PH2O = 0.40 atm
6. For the reaction below at 25°C, Kc = 115. If 0.050 moles of each reactant and 0.600 moles of
product are placed in a 1.00-L flask, what are the equilibrium concentrations?
H2(g) +
F2(g) 
2 HF(g)
0.050
+x
0.050+x
0.050
+x
0.050+x
0.600
–2x
0.600–2x
(0.600 – 2x)2 = 115 → square root both sides
(0.050 + x)2
→→ x = 0.0050 M
[H2] = [F2] = 0.055 M
[HF] = 0.590 M
Q = (0.600)2 = 144
(0.050)2