The generic reaction has the following rate laws: forward reaction: rate = k[A]^

Question

The generic reaction has the following rate laws: forward reaction: rate = k[A]^2 reverse reaction: rate = kT [B] where k_f is the rate constant for the forward reaction and k_T is the rate constant for the reverse reaction. At equilibrium, the two rates are equal and so k_f [A]^2 = k_r [B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions: K_c = [B]/[A]^2 = k_f/k_r Formation of nitrosyl bromide Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with bromine. Bro-2 2NO(g) + Br_2(g) 2NOBr(g) The reaction rapidly establishes equilibrium when the reactants are mixed. At a certain temperature the initial concentration of NO was 0.400 M and that of Bro was 0.265 M At equilibrium the concentration of NOBr was found to be 0.250 M What is the value of K_c at this temperature? Express your answer numerically.

Explanation / Answer

                   2 NO     +      Br2 ------> 2 NOBr

initially       0.4              0.265                0

at equi    (0.4 - 2x)     (0.265 - x)         2x

2x = 0.25 then x = 0.125

[NO] = 0.4 - 0.25 = 0.15

[Br2] = 0.265 - 0.125 = 0.14

Kc = [NOBr] ^2 / [NO]^2 [Br2]

Kc = (0.25)^2 / ((0.15)^2 * (0.14))

Kc = 19.8413

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.