Nitrosyl bromide, NOBr, readily dissociated according to the following equilibrium equation:
2 NOBr (g) ----> 2 NO(g) + Br2 (g)
NOBr was placed in a 1.00 L flask at 25.0oC and allowed to dissociated. At equilibrium, 0.17 moles of Br2 are
obtained. If NOBr is found to be 34.0% dissociated, what is the value of Kc?
I don't think the question can be answered; there is no quantity given for the initial NOBr. If 1 mol than the following:
..........NOBr ==> 2NO + Br2
E.......1-0.34.....0.34...0.17
Kc = (NO)^2(Br2)/(NOBr)^2
To find the value of Kc, we need to first determine the concentrations of the species at equilibrium. We are given that 0.17 moles of Br2 are obtained at equilibrium.
We also know that NOBr is 34.0% dissociated, which means that 34.0% of the initial moles of NOBr have dissociated into NO and Br2. This implies that 34.0% of the initial moles of NOBr have formed 0.17 moles of Br2.
Let's denote the initial moles of NOBr as x. Then, the moles of dissociated NOBr would be 0.34x, and the moles of Br2 obtained would also be 0.34x.
Since 2 moles of NOBr dissociate to form 2 moles of NO and 1 mole of Br2, we can write the concentration of Br2 at equilibrium as follows:
[Br2] = (0.34x)/1.00 L = 0.17 mol/L
Simplifying the equation, we find:
0.34x = 0.17
Solving for x, we get:
x = 0.17/0.34
x = 0.50 moles
Now, we can use this value of x to calculate the concentrations of NO and NOBr at equilibrium.
[NO] = (2 * 0.34x)/1.00 L = (2 * 0.34 * 0.50)/1.00 L = 0.34 mol/L
[NOBr] = (x - 0.34x)/1.00 L = (0.50 - (0.34 * 0.50))/1.00 L = 0.33 mol/L
Lastly, we can use these concentrations to calculate the value of Kc using the given equilibrium equation:
Kc = ([NO]^2 * [Br2])/([NOBr]^2)
= (0.34^2 * 0.17)/(0.33^2)
= 0.062
So, the value of Kc is 0.062.