At 473 K, for the elementary reaction
2NOCl <=> 2NO + Cl_2
k(1) =0.078 L/mol*s
k(-1)=470 L^2/mol^2*s
A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.49 mol/L. Find Cl_2 and NO concentrations
I just came here to find the answer
evry licuid + blue luqid
To find the concentrations of Cl_2 and NO at equilibrium, we can use the equation for the reaction quotient (Qc) and the equilibrium expression (Kc).
The equation for the reaction quotient (Qc) is given by:
Qc = [NO]^2 * [Cl_2] / [NOCl]^2
The equilibrium expression (Kc) is given by:
Kc = k(1) / k(-1)
At equilibrium, Qc is equal to Kc. Therefore, we can set up the following equation:
[NOCl]eq^2 = ([NO]eq)^2 * [Cl_2]eq
Substituting the given values:
(0.49)^2 = ([NO]eq)^2 * [Cl_2]eq
Now, we need to solve for [Cl_2]eq and [NO]eq.
First, let's solve for [Cl_2]eq:
[Cl_2]eq = (0.49)^2 / ([NO]eq)^2
Substituting the given value of [NO]eq as 0.49 mol/L, we get:
[Cl_2]eq = (0.49)^2 / (0.49)^2
Simplifying, we find:
[Cl_2]eq = 1 mol/L
Therefore, the concentration of Cl_2 at equilibrium is 1 mol/L.
Next, let's solve for [NO]eq:
Using the equation:
[NOCl]eq = [NO]eq + ([NO]eq)^2
Substituting the given value of [NOCl]eq as 0.49 mol/L, we get:
0.49 = [NO]eq + ([NO]eq)^2
Rearranging, we have a quadratic equation:
([NO]eq)^2 + [NO]eq - 0.49 = 0
Solving this quadratic equation using the quadratic formula, we get:
[NO]eq = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values a = 1, b = 1, and c = -0.49, we get:
[NO]eq = (-1 ± √(1^2 - 4*1*(-0.49))) / (2*1)
Simplifying further, we find:
[NO]eq = (-1 ± √(1 + 1.96)) / 2
[NO]eq = (-1 ± √2.96) / 2
Using the quadratic formula, we get two possible values for [NO]eq:
[NO]eq ≈ -0.76 mol/L or [NO]eq ≈ 0.76 mol/L
Since concentration cannot be negative, we can discard the negative value. Therefore, the concentration of NO at equilibrium is approximately 0.76 mol/L.
In summary, at equilibrium, the concentration of Cl_2 is 1 mol/L and the concentration of NO is 0.76 mol/L.
concentration of Cl2 will be 0.0084 mol/L
concentration of NO will be 0.017mol/L
This is because there are two moles of NO but only one mol of Cl2. We obtain the concentration of Cl2 by calculating the number of moles (from the ratio) and dividing this by the volume you calculate