Chemical Equilibrium

  • Thread starter future_vet
  • Start date
  • #1
169
0
Hello,

I have a few questions about my homework: (it is a rather long post, but I explained my work in detail, so everything is very clear -I hope-).

1) When we have a reaction such as: SO2 (g) + 1/2O2 (g) <=> SO3 (g) and then we have:
2SO2 (g) + O2 (g) <=> 2SO3 (g), we calculate Kc of the second reaction as = the first Kc squared. Correct? Why isn't it 2Kc? Is there a case where we would have Kc being twice as much?

2) We have the following reaction:
N2 (g) + 3H2 (g) <=> 2NH3 (g). Kp = 4.51 x 10^(-5) at 450 degrees celsius.
We are given values for each of the reactants and product.
105 atm NH3, 35 atm N2 and 495 atm H2.
We have to calculate whether the reaction is at equilibrium or not at 450.
Which I did by calculating Kp by plugging in the values, and then comparing my result to the value of Kp given at the begining of the problem. I got 1.20 x 10^(-3), which is not the same, and therefore the reaction is not at equilibrium. Correct?
But then, I have to indicate in which direction the mixture must shift in order to achieve equilibrium. I can't seem to figure out how I should do that.
Do I calculate Qp and then see which is bigger? (Qc bigger, it shift to the left, Qc smaller, it shifts to the right).

The thing is, I must be forgetting something, because I don't know which values to use to get Qp.

3) We have a mixture of 1.374g of H2 and 70.31 g of Br2, heated in a 2.00L vessel at 700K. We get this reaction:
H2 (g) + Br2 (g) <=> 2 HBr (g)
At equilibrium, we have H2 = 0.566g. We need to calculate the equilibrium concentrations of H2, Br2, HBr, and calculate Kc.
For some reason, I am blocking on this problem. Could you give me some pointers so I can start thinking in the right direction?

Thank you,

J.
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
future_vet said:
Hello,

I have a few questions about my homework: (it is a rather long post, but I explained my work in detail, so everything is very clear -I hope-).

1) When we have a reaction such as: SO2 (g) + 1/2O2 (g) <=> SO3 (g) and then we have:
2SO2 (g) + O2 (g) <=> 2SO3 (g), we calculate Kc of the second reaction as = the first Kc squared. Correct? Why isn't it 2Kc? Is there a case where we would have Kc being twice as much?
It follows directly from the definition of Kc, doesn't it? Write down the expression for Kc for the two equations, and you'll see that the second one is simply the square of the first.

Ther is no good reason to expect a doubling of the coefficients to result in a doubling of Kc (except in the case where Kc=2).

2) We have the following reaction:
N2 (g) + 3H2 (g) <=> 2NH3 (g). Kp = 4.51 x 10^(-5) at 450 degrees celsius.
We are given values for each of the reactants and product.
105 atm NH3, 35 atm N2 and 495 atm H2.
We have to calculate whether the reaction is at equilibrium or not at 450.
Which I did by calculating Kp by plugging in the values, and then comparing my result to the value of Kp given at the begining of the problem. I got 1.20 x 10^(-3), which is not the same, and therefore the reaction is not at equilibrium. Correct?
Show the calculation you did to get this number. I get a different number.

Also, the number you are calculating now is the Qp, not the Kp. You can only know that you are calculating Kp if you know that the pressures given represent equilibrium values.

3) We have a mixture of 1.374g of H2 and 70.31 g of Br2, heated in a 2.00L vessel at 700K. We get this reaction:
H2 (g) + Br2 (g) <=> 2 HBr (g)
At equilibrium, we have H2 = 0.566g. We need to calculate the equilibrium concentrations of H2, Br2, HBr, and calculate Kc.
For some reason, I am blocking on this problem. Could you give me some pointers so I can start thinking in the right direction?
Standard approach for all such problems:

1. Let the initial concentrations (which can easily be calculated in this problem, from the masses and the volume of the container), be [A] and .

2. The stoichiometry tells us how much of A reacts with how much of B. In this case, 1 mole of A reacts with 1 mole of B. So, in equilibrium, if x mol/L of A have been consumed, then x mol/L of B have been used up too. And again, from the stoichiometry, we know that 2x mol/L of C (the product) have been formed (only in this particular case, where A + B -> 2C ).

3. So, the equilibrium concentrations can be calculated. They are respectively: [A]-x, -x and 2x.

4. The equilibrium constant can then be evaluated. It is simply Kc = (2X)^2/(([A]-x)(-x))
 
  • #3
169
0
Calculating moles out of masses

Gokul43201 said:
1. Let the initial concentrations (which can easily be calculated in this problem, from the masses and the volume of the container), be [A] and .


I am not 100% certain how I should do that.
Do I find the molecular weight of each of the chemicals, and do, for H2 for example,
1.374 g H2 = 1.374 g x 1 mol / (1.008 x2) g = 0.682 mol H2 ?

Thank you,

J.
 
  • #4
169
0
Gokul43201 said:
Show the calculation you did to get this number. I get a different number.
I tried calculating again, and got the following:
Qp = (105^2) / (35 x 495^3) = 2.60 x 10^-6.
Would that be correct?

In order for the reaction to be at equilibrium, we need Qp = Kp, correct? In that case, Kp is larger. Qp < Kp => the reaction has to shift to the right to achieve equilibrium?

Thank you so much for you help!!
 
  • #5
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
future_vet said:
I am not 100% certain how I should do that.
Do I find the molecular weight of each of the chemicals, and do, for H2 for example,
1.374 g H2 = 1.374 g x 1 mol / (1.008 x2) g = 0.682 mol H2 ?
Exactly right! Then you divide the number of moles by the volume of the container (2L), to get the concentration in mol/L. First complete this step. Then, similarly calculating the value of the quilibrium concentration of H2 and subtracting from the initial concentration, you can find the change in concentration, x. Once you have x, you can find Kc.

I tried calculating again, and got the following:
Qp = (105^2) / (35 x 495^3) = 2.60 x 10^-6.
Would that be correct?
Yes, that looks more like the number I got.

In order for the reaction to be at equilibrium, we need Qp = Kp, correct? In that case, Kp is larger. Qp < Kp => the reaction has to shift to the right to achieve equilibrium?
This is correct as well.
 
  • #6
169
0
Thank you Gokul43201, you are extremely helpful!

~J.
 

Related Threads on Chemical Equilibrium

  • Last Post
Replies
2
Views
902
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
12
Views
7K
Replies
0
Views
8K
  • Last Post
Replies
5
Views
2K
Top