A 100 L reaction container is charged with 0.762 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction:
NOBr(g) ↔ NO(g) + 0.5Br2(g)
At equilibrium the bromine concentration is 1.56x10-3 M. Calculate Kc (in M0.5)
**Not specifying the temperature allows for a more liberal use of random numbers.
I have no idea what (in M0.5) stands for. M means mols/L but the 0.5 leaves me stumped. Anyway-------
You have 0.762 mols in 100 L so (NOBr) = 0.762 moles/100 L = 0.00762 M initially.
....................NOBr(g) ↔ NO(g) + 0.5Br2(g)
I.....................0.00762M.......0...........0
C.......................-x................+x.........0.5x
E...............0.00762 - x............x....... 0.5x
but the problem tells you that (Br2) = 0.00156 M = 0.5x
Kc = (NO)(Br2)^0.5/(NOBr)
Evaluate NOBr, NO and Br2 then
Substitute the E line into the Kc expression and solve for Kc. Post your work if you get stuck. I get an answer of 0.027 but it's late late and past my bed time. Be sure you check that out carefully.
I have no idea what (in M0.5) stands for. M means mols/L but the 0.5 leaves me stumped. Please help; perhaps I can work the problem if I know what that stands for. Thanks.
To calculate the equilibrium constant Kc (in M^0.5), we need to know the initial and equilibrium concentrations of each species involved in the reaction.
We are given that the initial moles of NOBr is 0.762 mol, and the reaction is carried out in a 100 L container. Therefore, the initial concentration of NOBr is:
[NOBr]initial = 0.762 mol / 100 L
= 0.00762 M
At equilibrium, we are given that the concentration of bromine ([Br2]) is 1.56x10^(-3) M.
Since we have a stoichiometric coefficient of 0.5 in front of Br2, we need to multiply its concentration by 2 to get the concentration of Br2 in the balanced equation:
[Br2]equilibrium = 2 * 1.56x10^(-3) M
= 3.12x10^(-3) M
To calculate the concentration of NO at equilibrium, we need to know its stoichiometric coefficient. Since there is no coefficient given in the equation, we assume it to be 1.
We can set up an ICE (Initial, Change, Equilibrium) table to calculate the equilibrium concentration of NO:
NOBr(g) ↔ NO(g) + 0.5Br2(g)
Initial (M) 0.00762 0 0
Change (M) -x +x +0.5x
Equilibrium (M) 0.00762-x x 0.5x
We can see that the decrease in NOBr concentration is equal to x. Therefore, the equilibrium concentration of NO ([NO]) is also equal to x.
We are given that [Br2]equilibrium = 3.12x10^(-3) M, which is equal to 0.5x. Therefore:
0.5x = 3.12x10^(-3)
x = (3.12x10^(-3)) / 0.5
x = 6.24x10^(-3) M
Now we have the equilibrium concentration of NO, which is 6.24x10^(-3) M.
Finally, we can calculate the equilibrium constant Kc using the concentrations of the species at equilibrium:
Kc = ([NO]^1 * [Br2]^0.5) / [NOBr]
= (6.24x10^(-3))^1 * (3.12x10^(-3))^0.5 / 0.00762
= 0.033 M^0.5
Therefore, the equilibrium constant Kc (in M^0.5) is 0.033.