Question

The reaction 2 NO2 ? 2 NO + O2 has a rate constant of k = 0.63 M-1 s-1.

(a) Based on the units of k, is this reaction first or second order?
(b) If the initial concentration of NO2 is 0.100 M, how would you determine how long it would take for the concentration to decrease to 0.025 M?

Answer 1

Answer:

Explanation:

Let's write out the equation again;

2 NO₂ →2 NO + O₂

We were given rate constant (K) = 0.63 M-1 s-1

a) How do we know if the reaction is a first or second order?

Let's write out the rate law for both order of reactions.

Order of reactions are pretty much the sum of the powers of the concentrations in a rate law.

First order

Rate = k[A]¹

Unit for concentration = M

Unit for Rate = M / t

Solving for K we have;

K = $\frac{Rate}{[A]} = \frac{M / t}{M} = \frac{1}{t}$ = s-1

Second order

Rate = k[A]²

Unit for concentration = M

Unit for Rate = M / t

Solving for K we have;

K = $\frac{Rate}{[A]²} = \frac{M / t}{M²} = \frac{1}{tM}$ = M-1 s-1

From this we can tell that the reaction is a second order reaction.

b) Initial Concentration (A₀) = 0.100M

Final concentration (A)= 0.025M

time?

The formular for this is given as;

$\frac{1}{A} = \frac{1}{A_0} + kt$

Solving for t, we have;

kt = $\frac{1}{A} - \frac{1}{A_0}$

0.63t = 1/0.025 - 1/0.1

0.63t = 40 - 10 = 30

t = 30/0.63 = 47.62 s

Answer 2

a. It is the second order.

b. The time period is 48 s.

Details required to determine the reaction and time period:

The reaction 2 NO2 ? 2 NO + O2 has a rate constant of k = 0.63 M-1 s-1.

Calculation of reaction and the time period:

(a)

The reaction is second order since k has units of $M^{-1}s-^{1}$

$Rate (M s^{-1}) = k (M^{-1}s^{-1}) \times [NO2]^2 (M^2)$

(b)

For a second-order reaction:

$1\div [NO2] = 1\div [NO2]o + kt$

Here

[NO2] is concentration at time t,

[NO2]o is initial concentration

$1\div 0.025 = 1\div 0.100 + 0.63 \times t$

Time t = 48 s

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