Supplemental Instruction
Iowa State University
Leader: Kelsey
Course: Chemistry 178
Instructor: Verkade
Date: 09/14/2010
1.
Given the following information:
𝐻𝐹 →
(𝑎𝑞) ←
𝐻
+
(𝑎𝑞)
𝐻
2
𝐶
2
𝑂
4(𝑎𝑞)
→
←
2𝐻 +
(𝑎𝑞)
+ 𝐹
−
(𝑎𝑞)
𝐾
𝐶
= 6.8 × 10 −4
+ 𝐶
2
𝑂 2−
4(𝑎𝑞)
𝐾
𝐶
= 3.8 × 10 −6 determine the value of K
C
for the reaction:
2𝐻𝐹
(𝑎𝑞)
+ 𝐶
2
𝑂
2−
4(𝑎𝑞)
→
←
2𝐹
−
(𝑎𝑞)
+ 𝐻
2
𝐶
2
𝑂
4(𝑎𝑞)
Kc = ((6.8 x 10^-4)^2)x(1/(3.8 x 10^-6)
2.
The density of Tin (Sn) is 7.287 g/cm
3
, what is the concentration (molarity, M)?
7.287 g/cm^3 = 7.287 g/ml (7.287 g/mL)*(1000 mL/1L)*(1 mol Sn/ 118.7 g Sn) = X M
3.
Write the equilibrium constant expression for K
C
for each of the following reactions: a.
𝐶𝑂
2(𝑔)
+ 𝐻
2(𝑔)
→
←
𝐶𝑂
(𝑔)
+ 𝐻
2
𝑂
(𝑙)
Kc=([CO]/([H2][CO2]} b.
𝑆𝑛𝑂
2(𝑠)
+ 2𝐶𝑂
(𝑔)
→
←
𝑆𝑛
(𝑠)
Kc=([CO2]^2)/([CO]^2)
+ 2𝐶𝑂
2(𝑔)
4.
A closed system initially containing 1.000 x 10
-3
M H
2
and 2.000 x 10
-3
M I
2
at 448 degrees C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10
-3
M. Calculate K
C
at 448 degrees C for the reaction taking place, which is:
𝐻
2(𝑔)
+ 𝐼
2(𝑔)
→
←
2𝐻𝐼
(𝑔)
Make ICE table
Kc=(Equilibrium Concentration of H2 * Equilibrium Concentration of I2)/(Equilibrium
Concentration of HI squared)
Supplemental Instruction
1060 Hixson-Lied Student Success Center

294-6624

www.si.iastate.edu

5.
At 448 degrees C the equilibrium constant K
C
for the reaction
𝐻
2(𝑔)
+ 𝐼
2(𝑔)
→
←
2𝐻𝐼
(𝑔) is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448 degrees C if we start with 2.0 x 10 -2 mol of HI, 1.0 x 10 -2 mol of H
2
, and 3.0 x 10 -2 mol of
I
2
in a 2.00-L container.
Reaction will proceed toward Products (Q<K)
6.
Consider the equilibrium 𝑁
2
𝑂
4(𝑔)
→
←
2𝑁𝑂
2(𝑔)
∆𝐻° = 58.0 𝑘𝐽
In which direction will the equilibrium shift when a) N
2
O
4
is added, b) NO
2
is removed, c) the total pressure is increased by addition of N
2(g)
, d) the volume is increased, e) the temperature is decreased? a) Right b) Right c) Left d) Right e) Right