1. Dinitrogen pentoxide, N 2 O 5 , decomposes by first order kinetics with a rat

Question

1. Dinitrogen pentoxide, N2O5, decomposes by first order kinetics with a rate constant of 3.7 x 10-5 s-1 at 298K. (a) What is the half-life in hours for the decomposition; (b) if [N2O5]0= 2.33 x 10-2 mol/L, what will be the concentration after 2.0 hr ; how much time will elapse in minutes before the N2O5 concentration decreases from 23.3 mmol/L to 17.6 mmol/L?

2. In the reaction, 2NO(g) + O2(g) ? NO2(g), when the concentration alone of NO was doubled, the rate increased by a factor of 4. When both the NO and O2 concentrations were increased by a factor of 2 the rate increased by a factor of 8. (a) What is the rate law for the reaction? (b) What is the order of the reaction?

3. The rate constant of the first order reaction, 2N2O(g) ? 2N2 + O2(g), is 0.38 s-1 at 1000 K and 0.87 s-1 at 1030 K. Calculate the activation energy for this reaction.

4. The rate constant of a reaction increases by a factor of 1000 in the presence of a catalyst at 25?C. The activation of the original pathway is 98 kJ/mol. What is the activation of the new pathway, all other factors being equal?

Explanation / Answer

1)

(a)

Rate constant for the reaction, k = 3.7 x 10-5 s-1

For a first order reaction, we have the following relation between rate constant and half life of reaction :

t1/2 = 0.693/k

So, half life, t1/2 = 0.693/(3.7 x 10-5 s-1) = 1.873*104 s = 5.202 hrs

(b)

For a first order kinetics, we have :

ln(X0/X) = k*t

Here,

X0 = initial concentration

X = concentration after time t

k = rate constant

For this case, X denotes N2O5

Putting X0 = 2.33 x 10-2 mol/L ,t = 2 hr = 7200s and solving we get :

X = 1.785 *10-2 mol/L

(c)Putting X0 = 23.3 mmol/L and X=17.6 mmol/L and solving we get :

t = 7852.5 s = 2.106 hrs

(2)

Lets assume the rate law as :

r = k*[NO]x*[O2]y ---(1)

Doubling the concentration of NO increased 'r' by a factor of 4

r' = k*[2NO]x*[O2]y ---(2)

Since r'/r = 4, so from (1) and (2) we get : x = 2

Doubling both increased r by factor of 8, thus

r'' = k*[2NO]x*[2O2]y ---(3)

r''/r = 8 and x=2, so from (1) and (3), we get :

y = 1

So rate law becomes :

r = k*[NO]2*[O2]1

Order of reaction = x+y = 3

(3)

We know :

k = A*e(-E/RT) ,where

k = rate constant

E = activation energy

A = Arrehnius constant

Thus, we get :

k1/k2 = e(E/RT2 - E/RT1) ,

Putting k1 = 0.38 s-1, k2 = 0.87 s-1, T1 = 1000K, T2 = 1030K and R = 8.314 Jmol-1K-1 , we get :

E = 236.44 kJ/mol

(4)

From the Arrehnius equation, we get :
k1/k2 = e(E2/RT - E1/RT),

Putting k1/k2 = 1/1000, E2 = 98 kJ/mol, T = 25'C = 298 K, we get :
E2 = 80.885 kJ/mol

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