8. What is the purpose for using catalysts in reactions? 9. Determine the rate e
ID: 1007687 • Letter: 8
Question
8. What is the purpose for using catalysts in reactions? 9. Determine the rate expression for the following data. AJo (mol/L) 0.100 0.100 0.200 Alo [Blo (mol/L) 0.100 0.200 0.200 Rate (mol/L-s) 0.18 0.36 1.44 10, For the data given in question #9, calculate the rate constant with its label. 11. Ammonia can be made to burn according to the following reaction: 4NH3 + 502 4N0 + 6H20 Suppose that at a certain moment of the reaction the ammonia is reacting at a rate of 0.24 mol/L-s. What is the rate at which the oxygen is reacting? 2. In question #11, what is the rate at which the water is being formed? What are the four factors that control the rate of chemical reactions?Explanation / Answer
8. The purpose of using catalysts in the reactions is to fasten the kinetics of the reactions. The reaction becomes faster by decreasing the activation energy of the molecules and the catalyst remains unchanged at the end of the reaction.
9. By comparing the 1st and 2nd run, we see that the rate of the reaction became double by doubling the concentration of B and keeping the concentration of A constant. So, rate is directly proportional to the concentration of B. By comparing the 2nd and 3rd run, we see that the rate of the reaction became 4 times by doubling the concentration of A and keeping the concentration of B constant. So, rate is directly proportional to the square of concentration of A. The rate law becomes:
r = k [A]2[B]
where r is rate of reaction and k is rate constant.
10. For above rate law, we put the values of 1st run.
r = k [A]2[B]
0.18 = k [0.100]2[0.100]
k = 0.18*103 = 180 L2/mol2.s = 180 M-2 s-1
11. Rate of disappearance of NH3 = 0.24 mol/L.s
Moles of NH3 reacting in reaction = 4 mol
rate of the reaction = 0.24/4 = 0.06 mol/L.s
Moles of O2 reacting in reaction = 5 mol
rate of disappearance of O2 = 5(rate of reaction) = 5(0.06) = 0.30 mol/L.s
Rate at which oxygen is reacting = 0.30 mol/L.s
12. Rate at which water is being formed =>
moles of H2O being formed = 6
Rate of appearance of H2O = 6(rate of reaction) = 6(0.06) = 0.36 mol/L.s
13. There are many factors which influence the rate of the reaction. Those are concentration of reactants, temperature, surface area, solvent and catalyst. All these affect the rate of reaction a lot. More the concentration of reactants, faster is the reaction. More the temperature, more are the collisions in the molecules, more is the rate. More is the surface area of the particles, more is the rate. The rate of the reaction depends on aqueous or organic solvent. If the catalyst is present in the reaction, the reaction becomes faster.
14. Rate of the reaction is directly proportional to the concentration of the reactant. If the concentration of the reactant id doubles, the rate of the reaction will also become twice. So, correct answer is a.
15. Rate law given : rate = k [ICl][H2]
rate constant = k = 1.63 *10-1 L/mol.s
[ICl] = 0.25 mol/L
[H2] = 0.50 mol/L
rate = 1.63 *10-1 (0.25)(0.50)
= 0.0204 mol/L.s = 2.04*10-2 mol/L.s
Rate = 2.04*10-2 mol/L.s
16. By comparing the 1st and 2nd run, we see that the rate of formation of NO became 4 times by doubling the concentration of NOCl. By comparing the 1st and 3rd run, we see that the rate of formation of NO became 9 times by increasing the concentration of NOCl by 3 times. So, the rate of formation of NO is becoming square of the concentration of NOCl. The rate law becomes:
rate = k [NOCl]2
17. The reaction
A --------> B+C
order is 2
So, rate law is : rate = k [A]2
[A]o = 0.100 M
The reaction is 20 % complete means A left is 80 %.
[A] = 80 % of 0.100 M = 0.80*0.100 = 0.080 M
Time = t = 40.0 min
rate constant = k
Integrated rate law for 2nd order reaction is:
1/[A] = [1/[A]o + kt
1/0.080 = 1/0.100 + k (40.0)
12.5 = 10 + 40k
2.5 = 40 k
k = 0.0625 = 6.25*10-2 M-1 s-1
rate constant = 6.25*10-2 M-1 s-1
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.