2 The following is a first order reaction The rate constant

2. The following is a first order reaction. The rate constant for this reaction is 0.543 sec. The initial concentration of NO2 is 4.50 The reaction is overall exothermic 2 NO2 (g) 2 NO (g) O2 (g) a. What is the rate law expression for this reaction? b. How many seconds does it take for 85.0% of the NO2 to decompose? c. What is the concentration of NO2 after 85 seconds?

Solution

We are given a first order reaction.

2 NO₂ (g) 2NO (g) + O₂ (g)

a.)

Rate law expression for given first order reaction is,

rate = k[NO₂]²

The rate only depends on the concentration of reactants only.

b.)

According to the integrated rate law, for a first order reaction change in the concentration of reactant is given by equation,

ln[A] = ln[A₀] - kt

where,

[A] = Final concentration of reactant (after time t)

[A₀] = Initial concentration of reactant

k = Rate constant

t= Time

Note: Here ln is the natural log

We are given,

Initial Concentration = 4.5 M

Final Concentration = 15% of 4.5 M = 0.675 M       (As 85 % of NO₂ got decomposed)

k = 0.543 s^-1

On substituting the given values in above equation we get

ln[0.675 M] = ln[4.50 M] - ( 0.543 s^-1 ) * (t)

ln[0.675 M] - ln[4.50 M] = - ( 0.543 s^-1 ) * (t)

-1.897 = - ( 0.543 s^-1) * (t)

t = 3.494 s

c.)

According to the integrated rate law, for a first order reaction change in the concentration of reactant is given by equation,

ln[A] = ln[A₀] - kt

[A₀] = 4.5 M

k = 0.543 s^-1

t= 85 s

On substituting the given values in above equation we get

ln[NO₂] = ln[4.50 ] - ( 0.543 s^-1 ) * (85 s)

ln[NO₂] = 1.504 - 46.155 = - 44.651

[NO₂] = 4.058 * 10^-20 M