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.

1) 2HgO(s) <-> 2Hg(l) + O2(g)

What is the form of the equilibrium constant, Kc, for the reaction ?

1. Kc = [O2]
2. Kc = [Hg]^2[O2]
3. Kc = ([Hg]^2[O2])/[HO]^2
4. Kc = [O2]/[HgO]^2
5. None is correct

2) PbO(s) + CO2(g) <-> PbCO3(s)
	P(species) means pressure of species
What is the form of the equilibrium constant ?

1. None is correct
2. Kp = 1/P(CO2)
3. Kc = [PbCO3]^2/[CO2][PbO]
4. Kp = P(PbCO3) / P(CO2)
5. Kp = [PbCO3]/([PbO] * P(CO2))

3) 2CO(g) + O2(g) <-> 2CO2, relationship between Kc & Kp is

1. Kp = KcRT
2. Kp = Kc(RT)^-1
3. Kp = Kc(RT)^-2
4. Kp = Kc
5. Kp = Kc(RT)^2

4) The equilibrium constant, Kp, for the following reaction is 280 at
150deg Celcius. Suppose that a quantity of IBr is placed in a closed
reaction vessel and the system is allowed to come to equilibrium at
150deg Celcius. When equilibrium is established, the pressure of IBr
is 0.200 atm. What is the pressure of I2 at equilibrium ?

I2(g) + Br2(g) <-> 2IBr(g) + 11.7kJ

1. 0.012
2. 0.096
3. none
4. 0.0168
5. 0.067

5) Kc = 0.040 for the system below at 450deg Celcius

PCl5(g) <-> PCl3(g) + Cl2(g)

What is Kp at 450 deg Celcius ?

1. 0.40
2. 0.64
3. 5.2 * 10^-2
4. 2.4
5. 6.7 * 10^-4

6 C) Given that Co2(g) reacts with C(s) via the reaction

C(s) + CO2(g) <-> 2CO(g)

and Kp = 1.90 atm, what is the equilibrium partial pressure of CO2 if
1.00 atm of CO2 is placed in a vessel wth PURE SOLID CARBON ? (No CO
initially)

1. 0.55atm
2. 0.85atm
3. 0.60atm
4. 0.43atm
5. 0.51atm

7) Kp = 0.073 for : N2 + 3H2 <-> 2NH3

In a container initial partial pressures of N2, H2 and NH3 are each
1atm. A catalyst is combined and the reaction proceeds to equilibrium.
What is the equilibrium PARTIAL PRESSURE of H2 ?

1. 1.2atm
2. 1.4atm
3. 1.6atm
3. 1.8atm
5. 2.0atm

8) 2SO2(g) + O2(g) <-> 2SO3(g) comes to an equilibrium. Then the total
pressure is decreased by increasing the volume of the container.
What will be the efect on the amount of SO3(g) present ?

1. an increase
2. cannot be answered without value of Delta H
3. doesnt change at all
4. cannot be answered without Kc and Kp values
5. a decrease

9) A 2.000 liter vessel is filled with 4.000 moles of So3 and 6.000
moles of O2. When the reaction
	2SO3(gas) <-> 2SO2(gas) + O2(gas)
comes to equilibrium a measurement shows that only 1.000 mole of SO3
remains.
	How many moles of O2 are in the vessel at equilibrium ?

1. none is correct
2. 3.750 moles
3. 7.000 moles
4. 7.500 moles
5. 12.000 moles

10) The equilibrium constant for thermal dissociation of F2

	F2(g) <-> 2F(g)

is 0.300.
	If initially 1.00 mol F2 is placed in a 1.00L container, which of the
folowing is the correct number of moles of F2 that have dissociated at
equilibrium ?

1. 0.548
2. 0.239
3. 0.176
4. 0.213
5. 0.130
6. 0.474
7. 0.956
8. 0.418

11) [N2O4]o and [NO2]o are the initial concentrations of N2O4 and NO2,
respectively. [N2O4] and [NO2] are the equilibrium concentrations.
	In the reaction

		N2O4(g) <--> 2NO2(g)

at equilibrium, which of the following relationship applies?

1. [N2O4] = 2([N2O4]o - [NO2])
2. [N2O4]o - [N2O4] = 2([NO2]o - [NO2])
3. [N2O4]o - [N2O4] = 2[NO2]
4. -([N2O4]o - [N2O4]) = [NO2]
5. -2([N2O4]o - [N2O4]) = [NO2]o - [NO2]

13) At 990deg C, Kc is 1.6 for the following reaction. If 4.0mol of
CO, 3.0mol of H2O, 2.0 mol of H2, and 1.0 mol of CO2 are placed in a
5.0 liter container and allowed to reach equilibrium at 990deg
Celcius, which response includes all of the following statements that
are correct and no others ?

	CO2(g) + H2(g) <--> H2O(g) + CO(g)

	I. The concentration of H2 will be greater than 0.40 mol/L
	II. The concentration of H2O will be less than 0.60 mol/L
	III. The concentration of CO will be less than 0.80 mol/L
	IV. The concentration of CO2 will be greater than 0.20mol/L

1. I and IV
2. I, II, and IV
3. I and IV
4. I and II
5. I, II, III, and IV
6. III and II
7. III

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.

---

Clarification of Question by jwheel-ga on 15 Oct 2003 20:31 PDT

Hello, is anyone answering this question ? It has been stock on
"locked" for a few hours now. Im just wondering whether the system did
something wrong and not allowing other researchers answer this
question. Thank you.

---

Request for Question Clarification by answerguru-ga on 15 Oct 2003 21:08 PDT

Hi there,

I've been working on your questions and have just noticed that you
decided to split the questions up (in line with the suggestion by
bobbie7-ga). I'm not sure if you still want me to answer this - please
let me know how you'd like to proceed.

answerguru-ga

---

Clarification of Question by jwheel-ga on 15 Oct 2003 23:02 PDT

Hello,

I did not split these problems, I split up a second set of problems.
They are nothing like this set, and the listing price is much less.

Thus please proceed with answering. Thank you very much for your
effort and your time. Much appreciated.

---

Clarification of Question by jwheel-ga on 16 Oct 2003 05:49 PDT

Hello,

Whoever is working on these questions, please post whichever answers
you already got as of now and you may continue with the rest later. Im
considering splitting THESE questions too because its taking so much
time. You will still get paid, so that shouldnt be a problem.

Thank you.

Hello -

1)  Equilibrium constants, as discussed, can be represented as:


     [C][D]    where [C][D] are the concentrations of products
k = --------   and [A][B] are the concentrations of reactants
     [A][B]

in this case: 2Hg)(s) <-> 2Hg(l) + O2(g)

the expression can be written as:


     [Hg]^2 * O2
k = --------------
      [HgO]^2     

this is answer 3, (modified) - ([Hg]^2[O2])/[HgO]^2


2) applying the same to PbO(s) + CO2(g) <-> PbCo3(s)

gives:


     [PbCO3]
k = -------------
    [PbO] * CO2

this is answer 5 [PbCo3]/([PbO]*P(CO2))


3) Kc and Kp are related by the number of molecules present.  Kc is
the usual equilibrium constant, Kp is the same given in

terms of partial pressure - the relation is Kp = Kc(RT)^N  where N is
the number of stoichiometric molecules.

in this case - 2CO(g) + O2(g) <-> 2CO2 the equation becomes:

     ( CO2*RT )                  CO2           1                1
Kp = -----------------------  = ---------- * -------  = Kc * ------
     ( CO*RT )^2 * O2*RT         CO^2*O2       RT^2            RT^2

so, answer 3 - Kp = Kc(RT)^-2

reference: http://info.hartwick.edu/envirsc/Courses/chem108/CH16-2new/sld001.htm

4) So, in this equation - I2(g) + Br2(g) <-> 2IBr(g) 

   we have 0.200 atm of IBr and an equilibrium constant of 280:

       (0.200 atm IBr)^2
280 = -------------------------        Note: x is the same for both I2
and Br2
       (x atm I2) * (x atm Br2)         because the stoichometry
(number from
                                        the equation is the same

so:  x^2 = (0.200)^2/280 
     x = 0.012    - there are 0.012 atmospheres of both I2 and Br2 at
equilibrium

answer - 1

5)  conversion of Kc to Kp  ->    Kp = Kc(RT)^delta n  - in this case
delta n is equal to 1 (the number of moles generated is

2, the number of moles to start is 1, therefore the delta (or change)
is 1), so Kp = Kc * RT.  R = 0.0821  T = 450 + 273 (T

is in Kelvin which is 273 lower than 0 celsius) - so:

Kp = 0.040 * 0.0821 * 723 = 2.4 - answer 4.

6)  solid things are neglected in equilibrium equations, so, the
equation is given as:

       CO^2                    CO^2
Kp = ----------- = 1.90 atm  = ----------- 
       CO2                     1.00


so:  CO^2 = 1.90
     CO = 1.4 atm - answer 2.

7)  In this case - N2 + 3H2 <-> 2NH3
      NH3^2
k = -----------  =  0.073
    N2 * H2^3

if each starts at 1 atm, the change to equilibrium is:

         (1 + 2x)^2        
0.073 = ------------- 
        (1-x)*(1-3x)^3      

solving for x gives -0.2 - so the partial pressures at equilibrium
are:
NH3 - 0.6
N2 -  1.2
H2 -  1.6

answer - 3

 
8)  The answer in this case can be determined from examining the
number of molecules of reactants versus the number of molecules of
product - 3 reactant (2 SO2 and 1 O2) vs. 2 product - dilution will
thus produce a greater effect upon the products, diminishing their
relative concentration - this will be compensated by a decrease in the
quantity of product.

answer 5 - SO3 will decrease


9)  2 SO3 (g) <-> 2SO2(g) + O2(g)
moles  4.000                6.000        at start
moles  1.000                 x           at equilibrium

as the ratio of SO3 to O2 is 2:1, a decrease of 3 moles of SO3 yields
1.5 moles of O2 - thus, at equilibrium, there will be 7.500 moles of
O2

answer - 4

10)  F2(g)     <->  2F(g)
moles 1.00                  at start
      1.00 - x      2x      at equilibrium

the equilibrium equation is:

       F^2                  (2x)^2
k = ----------  = 0.300 = ---------
       F2                  1.00 - x

x = 0.239  - note, this is the answer to the question, as it asks for
the number of moles which have dissociated, not for the concentration
of F2 at equilibrium - x represents the number of molecules which are
changed into F during the equilibration.

answer - 2

11) This question is closely related to #10 above, once more there is
one gas molecule dissociating into 2 gas molecules.  The change in the
concentration of N2O4 is  [N2O4]o  - [N2O4]
NO2  is  [NO2]o - [NO2]

but as there are twice as many molecules of NO2 produced per N2O4
consumed, the change in NO2 is twice as much as the change in N2O4,
and the direction of change must be opposite - therefore:

-2([N2O4]o - [N2O4]) = [NO2]o - [NO2] - answer 5


13)   CO2(g) + H2(g) <---->  H2O(g) + CO(g)
moles 1.0      2.0            3.0      4.0      at start
      1.0 + x  2.0 + x        3.0 - x  4.0 -x   at equilibrium


       H2O*CO          (3.0-x)(4.0-x)     x^2 - 2x + 12
k = ------------- = 1.6 = --------------- = ---------------
       H2*CO2          (1.0+x)(2.0+x)     x^2 + 3x + 2

so:  0.6x^2 + 6.8x - 8.8  and x = 1.2

moles 2.2     3.2             1.8     2.8    at equilibrium

So, only I and IV are correct, this is answer 3

14)   N2(g)  + 3H2(g)  <--->  2NH3
moles 4.50     8.00           4.00      at start
      

Plugging the numbers into the equilibrium equation:

      NH3^2             4.0^2
x = -------------  = --------------- = 0.027
     N2*H2^3         4.5*8.0^3

x is less than the k for this reaction which indicates that it will
shift towards the reactants.

answer 2 - more NH3 will be formed.

15) By reversing the reaction equation, it changes what is considered
the product and what is considered the reactant, so:

                    B^2
Kc = 8.68 * 10^5 = -----
                     A

              A
and Kc*  = ------
             B^2  

since this is the reciprocal equation, the Kc* must be the reciprocal
of Kc, thus: Kc* = 1.15 * 10^-6

16) HCHO(g) <---> H2(g) + CO(g)
moles  1.0       1.0      1.0


    H2 * CO      1.0 * 1.0
k = --------- = ----------- = 1.0  - answer 4
    HCHO            1.0


17) HCHO(g) <---> H2(g) + CO(g)

the ratio of reactant to product is 1:2
if 2.9 moles are added to the reactant, and 1.9 are added to the
product, this is in excess of the 1:2 ratio - this will shift the
reaction towards the reactants, forming more HCHO - answer 4


18) 4 NH3(g) + 5 O2(g)  <---> 4 NO(g)  + 6 H2O(g)



    NO^4 * H2O^6
k = --------------
    NH3^4 * O2^5

or (NO^4*H2O^6)/(NH3^4*O2^5) - answer 1

19)    A       +         B  <--->     C     +      2D
moles 2.0*10^-2      1.7*10^-4     2.4*10^-6     3.5 * 10^-3


     C * D^2       (2.4*10^-6)(3.5*10^-3)^2
x = ---------   =  ------------------------- =  7.2 * 10^-6
     A * B         (2.0*10^-2)(1.7*10^-4)


x < k = 3.7 * 10^-3  thus the forward reaction is favored - answer 2


Please let me know if you have any further questions or need
clarification for any of the answers.


synarchy


References available on the web:

A very large collection of sites devoted to explaining chemical
equilibrium, including several interactive applets:
http://www.chemistrycoach.com/tutorials-4.htm#Equilibrium

Another tutorial page on chemical equilibrium:
http://dbhs.wvusd.k12.ca.us/Equilibrium/Equilibrium.html


A few Java based applets that demonstrate the principals of chemical
equilibria:
http://bertrand.home.mindspring.com/chem/

A nice compilation chapter on chemical equilibrium:
http://www.chem1.com/acad/webtext/chemeq/

3 shockwave animations of equilibria:
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/equilvpBr2V8.html
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/CoCl2equilV8.html
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/no2n2o4equilV8.html

And another online chemistry textbook:
http://science.widener.edu/svb/tutorial/


search terms:
"general chemistry", chemistry, equilibrium

Thank you

Perhaps if you post each question separately at a lower price you may
have a better chance of getting this question answered.  It would be
faster this way also.

I just glanced at this and you should be careful, for example the
answer for the first question is incorrect as the chemical HgO has an
(s) next to it-indicating that it is a solid.  Solids excluded from
equilbrium calculations.