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Hey guys, here are a couple of gchem questions i cant seem to figure out. if anyone can help me, that would be great. thanx in advance:

1) if the activiation energy for a reaction in the forward direction is 96kJ, the activation energy for the reaction in reverse is 295 kj, and the energy of the products is 34kj, the energy of the reactants is?

a) - 233kj/mole
b) 233 kJ/mole
c) -199 kJ/mole
d) 199 kJ/mole
e) -34 kj/mole

answer: B

2) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above

answer: A
I picked (e) bc the molarity of PCL5 is 4.75m/l, and 40% of that is 1.896 which is the concentration of PCL3 and CL2. thus, when i put it into the equilibrium equ: Kc=(1.896)^2/4.75. this answer resembles that of A, but for some reason the denominator was divded in half. anyone know why?
 

dat_student

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Envision said:
Hey guys, here are a couple of gchem questions i cant seem to figure out. if anyone can help me, that would be great. thanx in advance:

1) if the activiation energy for a reaction in the forward direction is 96kJ, the activation energy for the reaction in reverse is 295 kj, and the energy of the products is 34kj, the energy of the reactants is?

a) - 233kj/mole
b) 233 kJ/mole
c) -199 kJ/mole
d) 199 kJ/mole
e) -34 kj/mole

answer: B

------329 KJ---------
|...........................|
|96 KJ....................|295KJ
|...........................|
[233 KJ reactant].....|
............................|
............................|
............................|
............................34 KJ (product)

34 KJ + (295 - 96) = 233 KJ
 

dat_student

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Envision said:
...

2) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above

answer: A
I picked (e) bc the molarity of PCL5 is 4.75m/l, and 40% of that is 1.896 which is the concentration of PCL3 and CL2. thus, when i put it into the equilibrium equ: Kc=(1.896)^2/4.75. this answer resembles that of A, but for some reason the denominator was divded in half. anyone know why?
only problem: you have 60% of 4.75 PCl5 left (i.e. 2.854). so you have to divide it by 2.85 not 4.75 :)
 
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allstardentist

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) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above

Kc= [PCl3][Cl2]/[PCl5]
= (14.25*.4/3)(14.25*.4/3)/(14.25*.6/3)
 

dat_student

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allstardentist said:
) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above

Kc= [PCl3][Cl2]/[PCl5]
= (14.25*.6/3)(14.25*.6/3)/(14.25*.4/3)
40% decomposes NOT 60%.

(14.25*.4/3)(14.25*.4/3)/(14.25*.6/3)
 
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Thanx, that clears up a lot.
 
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dat_student said:
if still ambiguities don't hesitate to ask. I hope you could decipher my drawing for the 1st Q
well, since you offered, do you think you can provide an explanation for the first question? i see what you did, im just not sure how you figured out how to do it that way. i suppose my understand of enthalpy is a bit limited, i'l probably have to go brush on that some more later. thanx, you've been a great help.
 

dat_student

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dat_student said:
------329 KJ---------
|...........................|
|96 KJ....................|295KJ
|...........................|
[233 KJ reactant].....|
............................|
............................|
............................|
............................34 KJ (product)

34 KJ + (295 - 96) = 233 KJ
Envision said:
well, since you offered, do you think you can provide an explanation for the first question? i see what you did, im just not sure how you figured out how to do it that way. i suppose my understand of enthalpy is a bit limited, i'l probably have to go brush on that some more later. thanx, you've been a great help.

energy of reactant + forward activation energy = energy of product + backward activation energy

energy of reactant + 96 KJ = 34 KJ + 295 KJ
energy of reactant = 34 + 295 - 96
energy of reactant = 233 KJ
 

dental#1

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2) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above
--------------------------------------------------------------------------
14.250/3= 4.75 * .40= 1.9 (decomposed)

14.250/3 = 4.75 * .60= 2.85

so.......... [PCl3][Cl2]/[PCl5] [1.9]^2/[2.85] answer choice a
 

keibee82

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dental#1 said:
2) When 14.250 moles of PCL5 gas is placed in a 3 liter container and comes to equilibrium at constant temperature, 40% of the PCL5 decomposes according to the equation:

PCl5 (g) <-> PCL3 (g) + Cl2(g)

what is the value of Kc for this reaction?

a) (1.896)^2/2.854
b) (2.854)^2/1.898
c) (3.8)^2/2.854
d) (2.854)(1.896)/3.8
e) none of the above
--------------------------------------------------------------------------
14.250/3= 4.75 * .40= 1.9 (decomposed)

14.250/3 = 4.75 * .60= 2.85

so.......... [PCl3][Cl2]/[PCl5] [1.9]^2/[2.85] answer choice a

can we use the calculator in the science section? or do we have to
do this by hand?
 
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keibee82 said:
can we use the calculator in the science section? or do we have to
do this by hand?
it can be done by hand if you know how to do it. for example, i can approximate 60% of 4.75 by lookin at the answer choices. thanx guys for your help.
 

slayerdeus

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dat_student said:
------329 KJ---------
|...........................|
|96 KJ....................|295KJ
|...........................|
[233 KJ reactant].....|
............................|
............................|
............................|
............................34 KJ (product)

34 KJ + (295 - 96) = 233 KJ
A great technique for this type of problem is drawing out the reaction path like dat_student did, it makes it so much easier to visualize the problem and what you should adding/subtracting.
 

bwestern

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keibee82 said:
can we use the calculator in the science section? or do we have to
do this by hand?
you have to do it by hand. its not that bad though. most of the gen chem answers are given like (0.1)/[(.05)(.2)]
so u really dont have to do many tough calculations, you just have to set up the equation correctly
 
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