The table below shows the initial rate of formation of O_2 for the following rea

Question

The table below shows the initial rate of formation of O_2 for the following reaction at 298K, at different concentrations of reactants. SO_2 + O_3 rightarrow SO_3 + O_2 (i) Using the data in the table, determine the rate equation for this reaction. (ii) Determine the rate constant and calculate the initial rate expected for experiment 4. (iii) Using the integrated form of the rate equation determined in (i) above, determine the t for experiment 1 and the amount of each reactant remaining after 10 seconds. (b) Discuss the effect of temperature on the rate of reaction. Your answer should include mathematical relationship, explanation of activation energy, and a brief overview of collision theory. (c) Experiment 1 in part (a) was earned out at 298K.. When it was repeated at 398K. the rate constant was determined to be 18.290mol^-1 Ls^-1. Determine the activation energy for this reaction.

Explanation / Answer

Q1.

a)

first, identify the rate equation

use initial rate methos

assume

Rate = k*[SO2]^a [O3]^b

when SO2 is constant ( experiment 1 and 2); O3 is halved, and the rate remains the same, so clearly O3 has no effect in rate, b = 0

then choose experiment 2 and 3

when we --> 0.750/0.254 --> 2.95, i.e. triple the concentration, then

rate --> (1.028)/(1.18*10^-10) = 8.7 --> i.e. 9

as we triple, the rate inreses by 3, so this must be to the second power, i.e. (3)^2 --> 9

Rate = k*[SO2]^2

b)

find K substitute any case

11.028 = k*(0.75^2)

K = 11.028 /(0.75^2) = 19.605

c)

Rate = 19.605 * [SO2]^2

solve for half time

1/(1/2*0.254) = 0.254 + 19.605*t

t = (1/(1/2*0.254) - 0.254 )/19.605 = 0.388 seconds

b)

the effect of T -- > increases rate since chaotic olecules will collide.

ln(K) = A*exp(-e/(RT))

ln(K) = A*exp(-EA/(8.314*298))

then, as T increases, k must increase as well

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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