Given the following rate law, how does the rate of reaction change if the concen

Question

Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? Rate = k [X][Y]^2 The rate of reaction will increases by a factor of 4. The rate of reaction will increases by a factor of 5. The rate of reaction will increases by a factor of 2. The rate of reaction will decreases by a factor of 2. The rate of reaction will remain unchanged. Determine the rate law and the value of k for the following reaction using the data provided. 2 NO(g) + O_2 rightarrow 2 NO_2(g) Rate = 3.1 Times 10^5 M^-3 s^-1 [NO]^2[O_2]^2 Rate = 1.7 Times 10^3 M^-2 s^-1 [NO]^2[O_2] Rate = 57 M^-1 s^-1 [NO][O_2] Rate = 9.4 Times 10^3 M^-2 s^-1 [NO][O_2]^2 Rate = 3.8 M^-1/2 3 s^-1 [NO][O_2]^1/2

Explanation / Answer

4. rate = k[X][Y]^2

If concentration of y is doubled , the rate of reaction increased by four times.

answer : A

5.

order.w.r.t NO

(0.03/0.06)^n = (8.55*10^-3)/(3.42*10^-2)

n = 2

order w.r.t O2

(0.0055/0.011)^n = (8.55*10^-3)/(1.71*10^-2)

n = 1

total order = 2+1 = 3

rate law , rate = k[NO]^2[O2]

8.55*10^-3 = K(0.03^2*0.0055)

K = 1727.3

Answer: B

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

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