In an experiment, 0.050 mol NO, 0.016 mol Cl2, and 0.50 mol NOCl are initially placed in a 4.0 L reaction vessel. The reaction that occurs is shown below. Assume Kc for this reaction is 2.2 x 10-5 . Please answer the following questions based on this information.
2 NOCl(g) = 2 NO(g) + Cl2(g)
a. Calculate the reaction quotient for this reaction.
b. Will the mixture form more NO or more NOCl or is the system at equilibrium? Explain your answer
reaction quotient and equilibrium?
2020-08-02 10:47 am
回答 (2)
2020-08-02 12:48 pm
✔ 最佳答案
Equilibrium quotient, Qc= [NO]² [Cl₂] / [NOCl]²
= (0.050/4)² (0.016/4) / (0.50/4)²
= 4.0 × 10⁻⁵
====
b.
As Qc > Kc, the equilibrium will shift to the left in order to lower the concentrations of NO and Cl₂ (the numerator of Qc) and increase the concentration of NOCl (the denominator of Qc). Hence, the mixture will form more NOCl.
2020-08-02 11:51 pm
Kc = [NO]^2 [Cl2] / [NOCl]^2
a. Initial concentrations:
[NO] = 0.0125 M
[Cl2] = 0.0040 M
[NOCl] = 0.125 M
Q = (0.0125)^2(0.0040)/(0.125)^2 = 4X10^-5
b. Q > Kc. Therefore, the mixture will form more NOCl and decrease the amounts of NO and Cl2, thus decreasing Q until Q = Kc.
a. Initial concentrations:
[NO] = 0.0125 M
[Cl2] = 0.0040 M
[NOCl] = 0.125 M
Q = (0.0125)^2(0.0040)/(0.125)^2 = 4X10^-5
b. Q > Kc. Therefore, the mixture will form more NOCl and decrease the amounts of NO and Cl2, thus decreasing Q until Q = Kc.
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