the equilibrium constant of the decomposition of NOCL is 1.6*10^-5. calculatethe concentration of all species at equilibrium if initially 3.52 mol of NOCL is placed in a 5.00 L container
You don't say if that is Kc or Kp; however, since you listed a volume I assume it is kc. M NOCl = 3.52/5.00 = about 0.7M but you can do it closer than that.
............2NOCl ==> 2NO + Cl2
I............0.70......0......0
C............-2x.......2x.....x
E............0.70-2x...2x.....x
Substitute and solve.
To calculate the concentration of each species at equilibrium, we need to use the equilibrium constant expression and the stoichiometry of the reaction. The balanced equation for the decomposition of NOCl is:
2 NOCl ⇌ N2 + Cl2
The equilibrium constant expression for this reaction is:
Kc = [N2] * [Cl2] / [NOCl]^2
Given that the equilibrium constant (Kc) is 1.6 * 10^-5 and the initial concentration of NOCl is 3.52 mol in a 5.00 L container, we can solve for the concentrations of N2 and Cl2 at equilibrium.
First, we need to determine the change in concentration of NOCl at equilibrium. Since the stoichiometry of the reaction is 2:1:1 (NOCl:N2:Cl2), the change in concentration of NOCl can be calculated as follows:
Change in [NOCl] = -2x (where x is the change in concentration of NOCl)
At equilibrium, the concentration of NOCl can be expressed as follows:
[NOCl] = Initial [NOCl] - 2x
We are given that the initial concentration of NOCl is 3.52 mol in a 5.00 L container. So, we have:
[NOCl] = 3.52 mol / 5.00 L = 0.704 M
Now, let's substitute these values into the equilibrium constant expression:
Kc = [N2] * [Cl2] / [NOCl]^2
1.6 * 10^-5 = [N2] * [Cl2] / (0.704)^2
Cross-multiplying:
1.6 * 10^-5 * (0.704)^2 = [N2] * [Cl2]
Finally, to solve for the concentrations of N2 and Cl2 at equilibrium, we need an additional piece of information, such as the initial concentration of N2 or Cl2, or the information about how the reaction progresses. Without that information, we won't be able to determine the specific concentrations at equilibrium.