Use the following information to determine the rate law: Exp 1: [A] = 1M, [B] =

Question

Use the following information to determine the rate law:

Exp 1: [A] = 1M, [B] = 1M, Rate = 1M/s

Exp 2: [A] = 1M, [B] = 2M, Rate = 1M/s

Exp 3: [A] = 2M, [B] = 1M, Rate = 4M/s

Rate=k[B]2

Use the following information to determine the rate law:

Exp 1: [A] = 1M, [B] = 1M, Rate = 1M/s

Exp 2: [A] = 1M, [B] = 2M, Rate = 1M/s

Exp 3: [A] = 2M, [B] = 1M, Rate = 2M/s

Rate=k[B]2

A first order reaction A--> B has an activation energy of 45 kJ/mol and a rate constant of 5 s-1 at 25

A. Rate = k[A] B. Rate=k[B] C. Rate=k[A][B] D. Rate=k[A]2[B] E. Rate=k[A][B]2 F. Rate=k[A]2[B]2 G. Rate=k[A]2 H.

Rate=k[B]2

Use the following information to determine the rate law:

Exp 1: [A] = 1M, [B] = 1M, Rate = 1M/s

Exp 2: [A] = 1M, [B] = 2M, Rate = 1M/s

Exp 3: [A] = 2M, [B] = 1M, Rate = 2M/s

A. Rate = k[A] B. Rate=k[B] C. Rate=k[A][B] D. Rate=k[A]2[B] E. Rate=k[A][B]2 F. Rate=k[A]2[B]2 G. Rate=k[A]2 H.

Rate=k[B]2

A first order reaction A--> B has an activation energy of 45 kJ/mol and a rate constant of 5 s-1 at 25

Explanation / Answer

1) rate = k[A]2

2) rate =  k[A]

3) as per Arrehenius equation:

ln(k2/k1) = (E/R)*{(1/T1) - (1/T2)} ; where k1 and k2 are the rate constants at temperatures T1 and T2 (T2>T1) and E = activation energy

thus, ln(k2/5) = (45000/8.314)*{(1/298) - (1/323)}

or, k2 = 20.4 s^-1

4) 4A + 4B --> 2C + D ; rate of reaction = 4

rate of the reaction = -(1/4)d[A]/dt

thus, rate of A = d[A]/dt = -16 M/s

5)  A --> B follows the rate law, Rate = 0.50 M-1s-1 [A]2.

since this is a seccond order reaction with rate constant,k = 0.5

thus, 1/[A] - 1/[A0] = k*t

or, t = {(1/80) - (1/100) }/0.5 = 0.00125 secconds

thus, the time to reduce 20% of the concentration is 0.00125 seconds

6) since the reaction is a first order

therefore, rate constant = 0.693/half life = 0.693/10 = 0.0693 s^-1

thus the correct option is (c)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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