2NO (g) + Cl2 (g) <=> 2NOCl (g) (reversible)
equilibrium constat = 1 x 10^7 @ certain temp
NO = 2.8 x 10^-2 mol
Cl2 = 1.4 x 10^-3 mol
NOCl = 5.9 mol
all mixed in a 1.4 L flask
what is Qc and what direction will system go?
(NO) = 2.8E-2/1.4L = about 0.02M
(Cl2) = 1.4E-3/1.4L = about 0.001M
(NOCl) = 5.9/1.4L = about 4.21
Qc = (NOCl)^2/(NO)^2*(Cl2)
Qc = (4.21)^2/(0.02)^2*(0.001) = approx 5E7 but you should confirm that. Compare that with Kc =1E7
Qc > Kc; therefore, numerator is too large and/or denominator is too small which means rxn must shift to the left in order to decrease numerator4 and increase denominator.
To determine the value of Qc, which is the reaction quotient, you need to write the expression for Qc using the molar concentrations of the species involved in the reaction.
For the given reaction 2NO (g) + Cl2 (g) <=> 2NOCl (g), the expression for Qc is:
Qc = [NOCl]^2 / ([NO]^2 * [Cl2])
Given the molar concentrations of the species:
[NO] = 2.8 x 10^-2 mol / 1.4 L = 2.0 x 10^-2 M
[Cl2] = 1.4 x 10^-3 mol / 1.4 L = 1.0 x 10^-3 M
[NOCl] = 5.9 mol / 1.4 L = 4.2 M
Plugging these values into the Qc expression:
Qc = (4.2)^2 / ((2.0 x 10^-2)^2 * (1.0 x 10^-3))
Calculating this value, Qc ≈ 882,000
Now, to determine the direction the system will go, compare the Qc value to the equilibrium constant (Kc).
Given equilibrium constant Kc = 1 x 10^7, we can see that Qc < Kc (882,000 < 1 x 10^7).
Since Qc is smaller than Kc, it suggests that the reactants are not fully consumed, and the system will proceed in the forward direction to reach equilibrium. Therefore, more NOCl will be formed, and the concentration of NO and Cl2 will decrease while NOCl concentration will increase.