Chemistry question on rate laws

Question

The rate constant for the decomposition of hydrogen iodide (HI) at 700oC equals 0.00160 M–1 .s–1 :

2HI(g) → H2(g) + I2(g)

1) Based on the units of the rate constant, is the reaction first order or second order? Explain.

2) Suppose a reaction is started with 0.0552 mol of HI in a volume of 735 cm3 . How many minutes will the reaction take for the moles of HI to drop by 66.6%?

3) Calculate the moles of HI that will remain after 15 minute and 25 seconds.

4) Calculate the half–life (in minutes) for the reaction.

Answer 1

(1) Let's look at the rate law and the units of k (the rate constant).

2HI(g) ==> H2(g) + I2(g)

Rate = k[HI]^m where m is the order of the reaction.

Now plug in the units: rate has units of concentration per unit time. In this case it would be M/s.

k has units of M^-1s^-1

M/s = M^-1s^-1[HI]^m

m = 2 so it is SECOND ORDER. In other words, M/s = M^-1s^-1 (M²)

(2) The integrated rate law for a second order reaction is 1/[A] = 1/[A]_o + kt

[A]o = 0.0552 mol/0.735 L = 0.0751 M

[A] = 0.334 x 0.0751 M = 0.0251 M (if it drops by 66.6%, 33.4% is left)

k = 0.00160 M^-1s^-1

t = ?

1/0.0251 = 1/0.0751 + 0.00160 t

39.8 = 13.3 + 0.00160 t

t = 16,563 sec x 1 min/60 sec = 276 minutes

(3) Use the same integrated rate law, and solve for [A] at time t (t in sec = 15 min x 60 sec//min = 900 + 25 = 925 sec.)

1/[A] = 1/0.0751 + (0.00160)(925)

1/[A] = 13.3 + 1.48 = 14.78

[A] = 0.0677 M

(4) The half life for a second order reaction is given as t_1/2 = 1/k[A]_o

t_1/2 = 1 / (0.0016)(0.0552)

t_1/2 = 11,322 sec x 1 min/60 sec = 189 minutes