CHEM126 – CHAPTER 14 – BASIC HOMEWORK / EQUILIBRIUM I

CHAPTER 14 BASIC HOMEWORK:

EQUILIBRIUM I

Question 1

Select one:
 Correct

Explanation

 CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 2

Question 2

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Correct

Explanation

 Look above!

Question 3

  CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 4
Select one:
 Correct
f. –

Explanation

 Look above!

* Remember NOT to include liquids and/or solids in the expression.

Question 4

  CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 5

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 Correct

Explanation

 Look above!

Again tricky tricky no solids and/or liquids.

Question 5

  CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 6

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 Correct

Explanation

 YOU GOT THIS KIDDOS

Question 6

Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g).
In a 5.00 L container the reactants and products are at equilibrium. There are 0.44 mol A, 0.30 mol B, 1.43 mol C, and 0.47 mol D. What is the equilibrium constant?

Help: Equilibrium constant

Additional hint in feedback

Correct

Explanation

Hint, remember to first write the equilibrium expression and convert the moles to concentrations.

K = [D]^2 / [A]^3
K= [0.47/5]^2/ [0.44/5]^3
K=12.966

Question 7

Consider the reaction: A(g) + 3B(s) ↔ 3C(g) + 2D(g).
In a 2.00 L container the reactants and products are at equilibrium. There are 0.47 mol A, 0.58 mol B, 1.84 mol C, and 0.27 mol D. What is the equilibrium constant?
Correct

Explanation

Hint, remember to first write the equilibrium expression and convert the moles to concentrations.
K = [C]^3[D]^2 / [A]
K = [1.84/2]^3[0.27/2]^2 / [0.47/2]
K=0.06038974

Question 8

  CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 7

Correct

Explanation

 2-x =1.14

x=0.86

y=3x=2.58

x=z=0.86

3H2 = 4 – 3(0.86) = 1.42

2NH3 = 0 + 0.86 = 0.86

Kc = [0.86]^2/ [1.14][1.42]^3= 0.145

Question 9

What is y, that is, how much hydrogen is used?

(Hint, use one of the methods used in the balance equation review problems to do the calculation. Enter amount as a positive quantity. An additional hint is given in the feedback.)

Correct

Explanation

Additional hint: What is x? It is the amount of nitrogen used. What is the relationship between the amount of nitrogen used and the amount of hydrogen used?

Question 10

What is the equilibrium number of moles of hydrogen?
Correct

 

Question 11

What is z, that is, how much ammonia is produce?

Correct

 

Question 12

What is the equilibrium constant, Keq?

(Remember to convert the number of moles to moles per liter or concentration in the equilibrium expression.)

Correct

 

Question 13

What is the equilibrium expression for:

Ca(OH)2(s) ↔ Ca2+(aq) + 2OH(aq)?

Select one:
 Correct

Explanation

IF YOU’RE NOT A GAS YOU CANT SIT WITH US

Question 14

Determine the equilibrium constant for the above reaction.

0.1 mol of Ca(OH)2 is placed in 2 L of water and stirred. At equilibrium, at a certain temperature, 0.0224 mol of Ca(OH)2 dissolves. What is Keq? (Assume, the volume of the solution is the same as the initial volume of water.)

Correct

Explanation

After determining the mol of Ca2+ and the mol of OH, remember to convert them to concentrations.
Kc= [Ca^2+][OH-]^2
(0.0224/2)(2*0.0224/2)^2 = 5.6e6

Question 15

Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g) Keq = 0.42 M-1.
[A] = 0.73 M and [D] =0.22 M. Some B and C are present. What is Q?

Help: Q and Keq

Correct

Explanation

 CHEM126 - CHAPTER 14 - BASIC HOMEWORK / EQUILIBRIUM I 8

 

Question 16

For the above value of Q
Select one:
 Correct

 

Question 17

Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g) Keq = 0.42 M-1

Same reaction and Keq, but this time:
[A] = 0.94 M and [D] =0.95 M. Some B and C are present. What is Q?

Correct

Explanation

 Q= [0.95]^2/[0.94]^3=1.09

Question 18

For the above value of Q

Select one:
 Correct

Explanation

 Q is large than Keq  so to obtain equilibrium the reaction must move to the left so there is more reactant and less product.