I worked the problem and obtained 2.440%
2NO ==> N2 + O2
initial:
NO = 6.745 M
N2 = 0
O2 = 0
change:
NO = -2x
O2 = +x
N2 = +x
equilibrium:
NO = 6.745-2x
O2 = x
N2 = x
Set up the ICE chart and solve for x, subtract 2x from the initial 6.745, divide the difference by 6.745 and convert to percent. Post your work if you want me to look for an error.
If the initial concentration of NO(g) is 6.745 mol/L, calculate the % of NO(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kc at 2400 K is 400.00. The initial concentration of the reaction products is 0 mol/L.
2NO(g) = N2(g)+O2(g)
The answer is 2.439.
3 answers
Thanks drbob, I was just getting confused by the direction of the equilibrium. how can you tell which way the equilibrium shifts?
There are two ways.
1. In this case, you start with zero O2 and zero N2 with 6.745 NO2. It has ONLY one way to go and that is to the right. Since there is nothing there to start with and lots of NO2, it can form only products. IF YOU HAD the same reaction and it starts with some O2 and some N2 but no NO2, it has only one way to go again and this time it goes to the left. So when there is nothing on one side, the decision is easy.
2. If you have something on BOTH sides, you must resort to a little chemistry. As an example, suppose we have the same reaction with initial as follows:
NO = 5
N2 = 2
O2 = 2
We write the reaction quotient, Q. Q is the same as K EXCEPT we put in concns we have (as opposed to K which is concns at equilibrium).
Kc = (N2)(O2)/(NO2)^2 = 400
Q = (2)(2)/(5^2) = 4/25 = 0.16. Now we compare this value of Q with Kc. Q is much smaller than Kc. Look at the fraction to see what this means. This means that the numerator is too small and the denominator is too large so the reaction must go in the direction to increase the products and decrease the reactants. That means to the right. Try another.
N2 = 5
O2 = 5
NO = 2
Q = (5)(5)/(2^2) = 6.25. Q < Kc so it goes to the right. Another.
N2 = 10
O2 = 10
NO = 0.001
Q = (10)(10)/(0.4)^2 = 625.
Q>Kc which means numerator is too large, denominator too large, reaction must go to the left.
1. In this case, you start with zero O2 and zero N2 with 6.745 NO2. It has ONLY one way to go and that is to the right. Since there is nothing there to start with and lots of NO2, it can form only products. IF YOU HAD the same reaction and it starts with some O2 and some N2 but no NO2, it has only one way to go again and this time it goes to the left. So when there is nothing on one side, the decision is easy.
2. If you have something on BOTH sides, you must resort to a little chemistry. As an example, suppose we have the same reaction with initial as follows:
NO = 5
N2 = 2
O2 = 2
We write the reaction quotient, Q. Q is the same as K EXCEPT we put in concns we have (as opposed to K which is concns at equilibrium).
Kc = (N2)(O2)/(NO2)^2 = 400
Q = (2)(2)/(5^2) = 4/25 = 0.16. Now we compare this value of Q with Kc. Q is much smaller than Kc. Look at the fraction to see what this means. This means that the numerator is too small and the denominator is too large so the reaction must go in the direction to increase the products and decrease the reactants. That means to the right. Try another.
N2 = 5
O2 = 5
NO = 2
Q = (5)(5)/(2^2) = 6.25. Q < Kc so it goes to the right. Another.
N2 = 10
O2 = 10
NO = 0.001
Q = (10)(10)/(0.4)^2 = 625.
Q>Kc which means numerator is too large, denominator too large, reaction must go to the left.
3 answers