# Homework 13 Solutions ```HW 13
Q1
What are the units of Kp and Kc for each of the following?
(a) NH4S(s)
Kp atm2
K c M2
(b) P4(g) + 5 O2(g)
Kp atm-6
Kc M-6
NH3(g) + H2S(g)
𝑲𝑲 = [𝑵𝑵𝑯𝑯𝟑𝟑 ][𝑯𝑯𝟐𝟐 𝑺𝑺]
𝑲𝑲 =
Q2
P4O10(s)
𝟏𝟏
[𝑷𝑷𝟒𝟒 ][𝑶𝑶𝟐𝟐 ]𝟓𝟓
Consider the following reaction:
2 H2S(g)
2 H2(g) + S2(g)
Kp was measured to be 0.0365 atm for this reaction at 385 K. What is the value of Kc at
385 K?
0.00115
Kc =
𝐾𝐾𝑐𝑐 =
M
�𝐾𝐾𝑝𝑝 �
0.0365
=
= 0.00115 𝑀𝑀
(𝑅𝑅𝑅𝑅)∆𝑛𝑛 𝑔𝑔 �0.0821(385)�1
What is the value for Kp for the following reaction at 385 K?
H2(g) + 1/2 S2(g)
Kp =
5.23
H2S(g)
atm-1/2
The reaction is reverse and half of the starting reaction, hence -1/2.
1
𝐾𝐾2 = 𝐾𝐾1−0.5 = (0.0365)−0.5 = 5.23 𝑎𝑎𝑎𝑎𝑚𝑚−2
Q3
What is the equilibrium concentration of Pb2+ ion in a solution of PbX2 that is 0.0420 M in X1ion?
PbX2(s)
[Pb2+] =
Pb2+(aq) + 2X1-(aq)
5.67e-04
M
K = 1.00e-06
𝐾𝐾 = [𝑋𝑋 −1 ]2 [𝑃𝑃𝑏𝑏 2+] → [𝑃𝑃𝑏𝑏 2+ ] =
𝐾𝐾
[𝑋𝑋 −1 ]2
=
1𝑥𝑥10−6
= 5.67𝑥𝑥10−4
(0.042)2
Q4
HX is a weak acid that reacts with water according to the following equation:
H3O1+(aq) + X1-(aq)
HX(aq) + H2O(l)
Ka = 2.51e-05
What is the equilibrium concentration of hydronium ion in a solution that is 0.0526 M in HX
and 0.253 M in X1- ion?
[H3O1+] =
𝑲𝑲𝒂𝒂 =
5.22e-06
M
[𝐻𝐻3 𝑂𝑂 +][𝑋𝑋 − ]
𝐾𝐾𝑎𝑎 [𝐻𝐻𝐻𝐻] 2.51𝑥𝑥10−5 (0.0526)
→ [𝐻𝐻3 𝑂𝑂 +] =
=
= 5.22𝑥𝑥10−6
[𝐻𝐻𝐻𝐻]
[𝑋𝑋 − ]
0.253
Q5
At some temperature, K = 16.2 for the following reaction:
H2(g) + CO2(g)
H2O(g) + CO(g)
The initial partial pressures of H2 and CO2 are each 1.90 atm.
What is the final pressure of CO?
1.52
PCO =
H2(g)
+
I
C
E
1.9
-x
19-x
𝐾𝐾 =
𝑥𝑥 2
= 16.2 →
(1.9 − 𝑥𝑥)2
atm
CO2(g)
1.9
-x
1.9-x
H2O(g)
+
0
+x
x
CO(g)
0
+x
x
𝑥𝑥 = 1.52 𝑎𝑎𝑎𝑎𝑎𝑎
What is the final total pressure?
Ptotal =
3.80
atm
Just add the starting partial pressures because the total P will not change. 2(1.9)= 3.8 atm
Q6
An equilibrium mixture contains 0.800 mol HI, 0.270 mol I2, and 0.150 mol H2 in a 1.00-L
flask. What is the equilibrium constant for the following reaction?
2HI(g)
Kc =
0.0633
H2(g) + I2(g)
𝐾𝐾 =
[𝐼𝐼2 ][𝐻𝐻2 ] 0.27(0.15)
=
= 0.0633
[𝐻𝐻𝐻𝐻]2
(0.8)2
How many moles of I2 must be removed in order to double the number of moles of H2 at
equilibrium?
0.367
mol I2
2HI(g)
H2(g)
+
E
0.8
0.27
0.15
I
0.8
0.27-x
0.15
C
-0.15
+0.15
+0.15
E
0.65
0.42-x
0.3
𝐾𝐾 =
[𝐼𝐼2 ][𝐻𝐻2 ]
[𝐻𝐻𝐻𝐻]2
→
(0.42−𝑥𝑥)(0.3)
(0.65)2
= 0.0633
→
I2(g)
𝑥𝑥 = 0.367 mol I2
Hint : If you remove I2, you have new &quot;initial&quot; concentrations. Think about which
of these you know, and which final concentrations you know.
```