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.
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.
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.
To calculate the percent of NO(g) left over after reaching equilibrium, you need to use the concept of equilibrium constants and calculate the quantities of reactants and products involved in the reaction.
The balanced equation for the reaction is: 2NO(g) = N2(g) + O2(g)
The equilibrium constant, Kc, is given as 400.00.
We know that initially, the concentration of NO(g) is 6.745 mol/L, and the concentration of the reaction products (N2 and O2) is 0 mol/L.
Let's assume that at equilibrium, the concentration of NO(g) remaining is x mol/L. Since we have a 2:1 stoichiometric ratio between NO and N2 in the balanced equation, the concentration of N2 at equilibrium will be 2x mol/L. Similarly, the concentration of O2 at equilibrium will also be 2x mol/L.
Now, using the equilibrium constant expression for the given reaction:
Kc = [N2(g)] * [O2(g)] / [NO(g)]^2
Plugging in the values:
400.00 = (2x) * (2x) / (6.745)^2
Simplifying the equation:
400.00 = 4x^2 / 45.451
Multiplying both sides by 45.451:
180.20 = 4x^2
Dividing both sides by 4:
45.05 = x^2
Taking the square root of both sides:
x ≈ 6.714
The concentration of NO(g) left over at equilibrium is approximately 6.714 mol/L.
To calculate the percent left over, use the formula:
% NO left over = (NO concentration at equilibrium / Initial NO concentration) * 100
% NO left over = (6.714 mol/L / 6.745 mol/L) * 100 ≈ 99.54%
Therefore, the percent of NO(g) left over after the reaction reaches equilibrium is approximately 99.54%, which rounds up to 99.5%.