Comments on: How do you calculate the equilibrium amounts of H2 and HF? http://vk3app.com/2010/02/how-do-you-calculate-the-equilibrium-amounts-of-h2-and-hf/ Latest technology news including developments in new technologies Sun, 01 Aug 2010 19:44:14 +0000 hourly 1 http://wordpress.org/?v=3.0 By: Dr.A http://vk3app.com/2010/02/how-do-you-calculate-the-equilibrium-amounts-of-h2-and-hf/comment-page-1/#comment-802 Dr.A Tue, 09 Feb 2010 02:18:50 +0000 http://vk3app.com/2010/02/how-do-you-calculate-the-equilibrium-amounts-of-h2-and-hf/#comment-802 initial concentration H2 = F2 = 1.00 / 10.0 = 0.100 M at equilibrium [H2]= [F2]= 0.100-x [HF]= 2x 1 x 10^2 = (2x)^2 / (0.100-x)(0.100-x) square root 10 = 2x/ 0.100-x 1 - 10 x = 2x 12 x =1 x = 0.0833 M [H2] = [F2]= 0.100 - 0.0833 =0.0167 M moles H2 = moles F2 = 0.0167 M x 10.0 L = 0.167 2x = 0.0833 x 2 =0.167 moles HF = 0.167 x 10.0= 1.67 initial concentration H2 = F2 = 1.00 / 10.0 = 0.100 M
at equilibrium [H2]= [F2]= 0.100-x
[HF]= 2x

1 x 10^2 = (2x)^2 / (0.100-x)(0.100-x)

square root
10 = 2x/ 0.100-x
1 – 10 x = 2x
12 x =1
x = 0.0833 M

[H2] = [F2]= 0.100 – 0.0833 =0.0167 M
moles H2 = moles F2 = 0.0167 M x 10.0 L = 0.167

2x = 0.0833 x 2 =0.167
moles HF = 0.167 x 10.0= 1.67

]]> By: Sara http://vk3app.com/2010/02/how-do-you-calculate-the-equilibrium-amounts-of-h2-and-hf/comment-page-1/#comment-801 Sara Tue, 09 Feb 2010 01:25:37 +0000 http://vk3app.com/2010/02/how-do-you-calculate-the-equilibrium-amounts-of-h2-and-hf/#comment-801 You can set up something that can be called an ICE table. I stands for initial, C stands for change, and E stands for equilibrium. All the amounts are calculated in Molarity. So you have the equation: H2(g) + F2(g) --> 2HF(g) So you can set up the table in the following way: Molarity:____[H2]_____[F2]_____[HF] I _________ .1 _____ .1________0 C__________- X______- X______+ 2X E_________.1 - X____.1 - X______2X For H2 and F2 you know that it's minus X because the initial amount is decreasing. Also, the coefficient on the X is the same as the coefficient in the balanced chemical equation. Now, the equation for the equilibrium constant, K, for the generic chemical equation aA + bB --> cC + dD is: K = ([C]^c[D]^d) / ([A]^a[B]^b) where the square brackets mean Molarity. So now substitute in your values. 1.0 x 10^2 = [2X]^2 / [.1 - X] [.1 - X] Now you could solve this in two ways. One is to use the quadratic formula to solve for X, and then substitute the value in to the ICE table to get your values, or you could assume X is small, and can be discounted. This means you can throw out X in any term where it is being added or subtracted, but no other time. Solving also, X can never be negative, so if you're solving the quadratic formula and you get a negative value, just throw it out and use the other one. You can set up something that can be called an ICE table. I stands for initial, C stands for change, and E stands for equilibrium. All the amounts are calculated in Molarity.
So you have the equation: H2(g) + F2(g) –> 2HF(g)
So you can set up the table in the following way:

Molarity:____[H2]_____[F2]_____[HF]
I _________ .1 _____ .1________0
C__________- X______- X______+ 2X
E_________.1 – X____.1 – X______2X

For H2 and F2 you know that it’s minus X because the initial amount is decreasing. Also, the coefficient on the X is the same as the coefficient in the balanced chemical equation.

Now, the equation for the equilibrium constant, K, for the generic chemical equation aA + bB –> cC + dD is:

K = ([C]^c[D]^d) / ([A]^a[B]^b)
where the square brackets mean Molarity. So now substitute in your values.

1.0 x 10^2 = [2X]^2 / [.1 - X] [.1 - X]
Now you could solve this in two ways. One is to use the quadratic formula to solve for X, and then substitute the value in to the ICE table to get your values, or you could assume X is small, and can be discounted. This means you can throw out X in any term where it is being added or subtracted, but no other time. Solving also, X can never be negative, so if you’re solving the quadratic formula and you get a negative value, just throw it out and use the other one.

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