To make a long story short, my youngest daughter decided to endure
through 4 semesters of chemistry, and now is paying up for it. She is
a talented little girl, but science is not her thing and it hurts for
me to see her study so hard but understand so little. It runs in the
family. I wanted to help myself, but me and my wife are no scientists
at all, plus it was so long ago. Tutoring is very very weak at where
she goes, and we know few friends around here who are capable of
helping, most of which already are looking at us the wrong way for
asking so much. I decided to open my pocket and let you, Chemistry
folks, help us out. I secretly took the next set of questions from my
daughters assignment section, and here they are. I understand that
many (or most) of you would have the problem to giving direct help
like this, "learning and understanding the material is the most
important", right ? Wrong. Only but a few of you would understand the
pain I feel when I see my daughter quietly cry in her room over her
low assignment grades after hours and hours of studying daily.
 
Below I copied down some of the questions and choice answers that came
along. I tried my best to copy down everything number to number, word
for word. Its a little bit difficult to transfer everything to text,
but I tried. Please feel free to clarify if you did not understand a
part.
 
I am asking for a list of answers to the problems, in numerical +
content format, as such :
 
Question 1: Answer 3. The number of moles is 4.0. 
Question 2: Answer 5. The reaction is reversed. 
 
Time is an issue, but I would like the answers to be correct. I do not
want to check my daughters answers and tell her that one of her
correct answers is incorrect simply because what was given to me is a
wrong answer. Once again I apologize for such rude question, but I am
desperate, and I ran out of all other options. I am also considering a
big tip for those who'll work hard on these, do all (or most) of the
questions, provide the correct answers and hopefully as soon as
possible.
 
Thank you once more. You are my last hope.


14) The reaction : N2(g) + 3H2(g) <-> 2NH3 
has an equilibrium constant Kc = 0.50 at 467deg Celcius. 
 What will happen eventually if 4.50 moles of N2, 8.00 moles of H2,
and 4.00 moles of NH3 are put in a 10-liter vessel and held at 467deg
Celcius ?
 
1. Nothing 
2. More NH3 will be formed 
3. More N2 and H2 will be formed 
4. 2NH3 will escape 
 
15) If Kc=8.68 * 10^5 for the reaction 
 A(g) <--> 2B (g) 
    What is Kc for the reaction written as: 2B(g) <--> A(g) 
 
<No Choices, Numerical Answer> 
 
16) Consider the following reaction. 
 
 HCHO(g) <--> H2(g) + CO(g) 
 
 1.0 mol of HCHO, 1.0 mol of H2 and 1.0 mol of CO exist in equilibrium
in a 2.0L reaction vessel at 600deg Celcius. Determine the value of th
equilibrium constant Kc for this system.
 
1. Impossible to determine 
2. None of these 
3. 0.50 
4. 1.0 
5. 2.0 
 
17) Consider the following reaction. 
 
 HCHO(g) <--> H2(g) + CO(g) 
 
2,9 moles if GCGI abd 1,9 mol if CI are theb added to this system.
Which of the following is now true ?
 
 
1. The reaction mixture remains at equilibrium. 
 
2. The reaction mixture is not at equilibrium, but will move towards
equilibrium by using up more HCHO.
 
3. The reaction mixture is not at equilibrium, but no further reaction
will occur.
 
4. The reaction mixture is not at equilibrium, but will move toward
equilibrium by forming more HCHO.
 
5. The forward rate of this reaction is the same as the reverse rate
at these new concentrations.
 
 
18) The expresion for Kc for the reaction at equilibrium is 
 
 4NH3(g) + 5O2(g) <--> 4NO(g) + 6H2O(g) 
 
([NO]^4[H2O]^6)/([NH3]^4[O2]^5) 
[NO]^4[H2O]^6 
[NH3]^4[O2]^5 
([NH3]^4[O2]^5)/([NO]^4[H2O]^6] 
 
19) Reaction : A + B <--> C + 2D 
 
This reaction has an equilibrium constant of 3.7 * 10^-3. Consider a
reaction mixture with
 
 [A] = 2.0 * 10^-2 M 
 [B] = 1.7 * 10^-4 M 
 [C] = 2.4 * 10^-6 M 
 [D] = 3.5 * 10^-3 M 
 
Which of the following is true ? 
 
1. The reverse reaction can occur to a greater extent than the forward
reaction until equilibrium is established.
 
2. The forward reaction can occur to a greater extent than the reverse
reaction until equilibrium is established.
 
3. No conclusions about the system can be made without additional
information.
 
4. The system is at equilibrium. 
 
5. Heat will be evolved.

Hi jwheel!!

Question 14: Answer 3. More N2 and H2 will be formed.

In effect Kc = [NH3]^2 / [N2]*[H2]^3 and equilibrium constants do not
change if you change the concentrations of things present in the
equilibrium. The only thing that changes an equilibrium constant is a
change of temperature.
If the conditions of the experiment have certain concentrations you
must calculate the number Q = [NH3]^2 / [N2]*[H2]^3 (= 0.694 in this
case).
If Q > Kc, then the system equilibrium will be shifted in order to do
Q =Kc; in this case Q must be decreased, so [NH3] must be decreased
and [N2] and [H2] must be increased.
Note about notation: The square brackets are concentration in
mol/liter.

-------------------------------------------------------------

Question 15: Answer: 1/(8.68 * 10^5) = 1.152 * 10^(-6)

By definition Kc of the reaction 2B(g) <--> A(g) is [A]/[B]^2
And Kc for the reaction  A(g) <--> 2B (g) is [B]^2/[A].
Then one Kc is the inverse of the other.

--------------------------------------------------------------

Question 16: Answer 3. Kc = 0.50  

For the reaction 
HCHO(g) <--> H2(g) + CO(g) 

Kc = [H2]*[CO]/[HCHO] (concentrations in mol/liter)

The concentrations are: 
[H2] = 1 mol / 2 liters = 0.5 m/l
[CO] = 1 mol / 2 liters = 0.5 m/l
[HCHO] = 1 mol / 2 liters = 0.5 m/l
Then
Kc = 0.5 * 0.5 / 0.5 = 0.5

-----------------------------------------------------------
Question 17: I think that there is a typo and I rewrite the question
as follows:
17) Consider the following reaction.  
  
 HCHO(g) <--> H2(g) + CO(g)  
  
2,9 moles of HCHO and 1,9 mol of CO are then added to this system. 
Which of the following is now true ? 



Question 17: Answer 2. The reaction mixture is not at equilibrium, but
will move towards equilibrium by using up more HCHO.
 

Kc = [H2]*[CO]/[HCHO] , and now we have more [HCHO] and [CO], in order
to find the equilibrium [HCHO] must be decreased (by using HCHO) and
[CO] and [H2] must be increased (and the only way is by using HCHO).

-----------------------------------------------------------

Question 18: Answer ([NO]^4.[H2O]^6)/([NH3]^4.[O2]^5)

By definition for the reaction A + 2B <--> 3C + 4D 
Kc = [C]^3 * [D]^4 / [A] * [B]^2

------------------------------------------------------------

Question 19: Answer 2. The forward reaction can occur to a greater
extent than the reverse reaction until equilibrium is established.

Q = [C]*[D]^2/[A]*[B] = 8.65 * 10^-6 < Kc .

This means that, in order to reach equilibrium, [C] and [D] must be
increased and [A] and [B] must be decreased or in other words more A
and B will be used to form more C and D.

-------------------------------------------------------------

I hope this helps you.
Please feel free to request any clarification needed before rate this
answer, I will gladly respond your requests.

Best regards.
livioflores-ga

Thanks, you are great