The reaction 2NO(g) + 02(g) rightarrow 2NO_2(g) was studied at 25 degree C, with

Question

The reaction 2NO(g) + 02(g) rightarrow 2NO_2(g) was studied at 25 degree C, with the following results: (a) Determine the rate law for the reaction and calculate the experimental rate constant. (b) Which mechanism (A or B) is consistent with the experimental rate law? Show how and find an expression for the experimental rate constant. (c) Sketch the energy profile versus the reaction coordinate. Assume Step 1 to be endothermic and the overall process to be exothermic. Clearly indicate in the graph the activation energies for the forward and reverse reactions. (d) The overall activation energy for this reaction is 45 kJ/mol. Calculate the rate constant at 45 degree C.

Explanation / Answer

Q2.

a)

Use the Method of Initial rates, which will allow us to determine the order:

recall that the rate of reaction could be modeled as

r = k [A]a [B]b

then, if we choose any points, i.e. 1 and 2 and divide them

r1/ r2 = (k [A]1^a [B]1^b) / (k [A]2^a [B]2^b)

If we were to choose any two points in which the initial rate are the same, we could cancel both concentrations and get a "a" and "b" values

so:

choose point 1 and 2

r1/ r2 = (k [A]1^a [B]1^b) / (k [A]2^a [B]2^b)

k constants cancel so

(1.4*10^-2)/(5.6*10^-2) = (0.012)^a (0.012)^b) / (0.024)^a (0.012)^b

then

0.25 = (0.012/0.024)^a

a = 2

it is 2nd order with respect to a

for b, choose points 2 and 3

r3/ r2 = (k [A]3^a [B]3^b) / (k [A]2^a [B]2^b)

(1.12*10^-1)/(5.6*10^-2) = (0.024)^a (0.024)^b /((0.024^a)(0.012^b))

2 = (0.024/0.012)^b

b = 1

it is of first order with respec tto b

so

overal order:

n = a + b = 2+1 = 3

it is 3rd order reaction

for k:

rate = k*[A]^2[B]

choose any point

1.4*10^-2 = k*(0.012^2)(0.012)

k = (1.4*10^-2 ) /((0.012^2)(0.012)) = 8101.85185 = 8.1*10^3

k = 8.1*10^3

b)

in order to be consisten, the elementary laws must be including, 2 times the A specie and 1 time the B specie

so:

A = NO

B = O2

The mechanism in B is the most probable

since we care on "SLOW" rates, those are the most predominant steps in the reaction

in B, we have two times the "NO" specie which marks the [A]^2 or [NO]^2

then, we have only 1 time the [B] specie, which is [O2]

then, the second mechanism is most likely to occur.

c.

If endothermic --> energy of products is higher than energy of reactants

If exothermic --> products are releasing heat, menaing the reactants had higher energy

d.

If = E = 45 kJ/mol = 45000 J/mol

then

T = 45C = 45+273 = 318K

from previous data:

k = 8.1*10^3 at T = 25C = 298 K

find k2

with help of ahrrenius equation

ln(k2/k1) = E/R*(1/T1 - 1/T2)

ln(k2/(8.1*10^3)) = 45000/(8.314)*(1/(298) - 1/(318))

solve for k2

ln(k2/(8.1*10^3)) = 1.1423

k2/(8.1*10^3) = exp(1.1423)

k2 = (8.1*10^3)*exp(1.1423)

k2 = 25385.14 = 2.5*10^4

makes sense since Temperature is increasing

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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