PLEASE HELP PROBLEMS 1-11.. PLEASE SHOW ALL STEPS AND WORK..!! THANK U..!! .1. F
ID: 889563 • Letter: P
Question
PLEASE HELP PROBLEMS 1-11.. PLEASE SHOW ALL STEPS AND WORK..!! THANK U..!!
.1. For the reaction given below, what is the instantaneous rate for each of the reactants and products?
3 A + 2 B 4 C
2.
. Given the following experimental data, find the rate law and the rate constant for the reaction:
NO (g) + NO2 (g) + O2 (g) N2O5 (g)
Run
[NO]o
[NO2]o
[O2]o
Initial Rate, M?s-1
1
0.10 M
0.10 M
0.10 M
2.1 x 10-2
2
0.20 M
0.10 M
0.10 M
4.2 x 10-2
3
0.20 M
0.30 M
0.20 M
1.26 x 10-1
4
0.10 M
0.10 M
0.20 M
2.1 x 10-2
3.
The half-life of a radioisotope is found to be 4.55 minutes. If the decay follows first order kinetics, what percentage of isotope will remain after 2.00 hours?
4.
The mechanism of a reaction is shown below.
HOOH + I¯ HOI + OH¯ (slow)
HOI + I¯ I2 + OH¯ (fast)
2OH¯ + 2H3O+ 4 H2O (fast)
a) What is the overall reaction?
b) Which compounds are intermediates?
c) Predict the rate law based on this mechanism.
d) What is the overall order of the reaction?
5.
For the reaction A + B C, the rate constant at 215 oC is 5.0 x 10-3 and the rate constant at 452o C is 1.2 x 10-1.
a) What is the activation energy in kJ/mol?
b) What is the rate constant at 100o C.
6.
The following reaction was studied at 656 K, and the data below were obtained:
2 NO2(g) ? 2 NO(g) + O2(g)
Time (min)
[NO2]
5.00
0.0142
10.00
0.0083
15.00
0.0058
20.00
0.0045
25.00
0.0037
30.00
0.0031
a) Based on the following graphs (next page) what is the order of the reaction?
b) What is the rate law?
c) What was the initial concentration of NO2?
d) What is the rate constant?
e) What is the half life?
7.
. The following reaction was studied in three separate experiments, and the data below were obtained:
2 NO (g) + O2 (g) ? 2 NO2 (g)
Experiment
[NO]
[O2]
Initial Rate (M/s)
1
0.0126
0.0125
1.41 x 10-2
2
0.0252
0.0125
5.64 x 10-2
3
0.0252
0.0250
1.13 x 10-1
a) What is the rate law for the reaction?
b) What is the order of the reaction?
c) What is the rate constant for the reaction?
d) What are the units of the rate constant?
8.
The decomposition of hydrogen bromide on a gold catalyst at 120 oC is zero order with respect to HBr. The rate constant is 7.40 x 10-5 M/s.
2 HBr(g) ? H2(g) + Br2(g)
a) Write the rate law for this reaction.
b) If the initial concentration of HBr is 0.870 M, calculate [HBr] after 45 minutes.
9.
The following two step mechanism has been proposed for the reaction:
Overall: 2 NO(g) + H2(g) ? N2O(g) + H2O(g)
1st step: NO(g) + NO(g) ? N2O2(g)
2nd step: N2O2(g) + H2(g) ? N2O(g) + H2O(g)
a) Show that the elementary steps of the proposed mechanism add up to provide the balanced equation for the reaction.
b) Write a rate law for each step of the reaction.
1st step:
2nd step:
c) List all intermediates in the reaction.
d) If the first step was known to be much slower than the second step, what is the rate law for the reaction?
10.
11.
A certain first-order reaction is 45.0% complete in 65 seconds. Calculate the rate
constant and the half-life for this process?
Run
[NO]o
[NO2]o
[O2]o
Initial Rate, M?s-1
1
0.10 M
0.10 M
0.10 M
2.1 x 10-2
2
0.20 M
0.10 M
0.10 M
4.2 x 10-2
3
0.20 M
0.30 M
0.20 M
1.26 x 10-1
4
0.10 M
0.10 M
0.20 M
2.1 x 10-2
Chapters 14 and 15 worksheet.pdf - Adobe Reader File Edit View Window Help My Files 8 | / 8 -+ | 9096 90% | Tools Sign Comment Graphs for Problem 1 Sign In Export PDF 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 Adobe ExportPDF Convert PDF files to Word or Excel online Select PDF File: 0.0571x Chapters 14 and 15 worksheet 1 file/127 KB Convert To: Microsoft Word (.docx) 0 51015 2025 30 35 Recognize Text in English(U.S.) Change t (min) Convert -4.0 4.2 4.4 -4.6 -4.8 Create PDF Send Files Store Files 5.2 5.4 5.6 5.8 6.0 0 10 15 20 25 30 35 t (min) 350 300 250 200 150 100 50 0 10.057x20.402 0 10 15 20 25 30 35 t (min) 8:01 AM 9/9/2014Explanation / Answer
Answer:
Given reaction:
3A +2B 4C
1). Rate (R) = -1/3 d(A)/dt = -1/2 d(B)/dt = 1/4 d(C)/dt
3).Half life (t1/2 = 4.55 minutes
t1/2=0.693/K
K = 0.693/4.55
=0.1523
K = 2.303/t[log a/a-x]
t = 2 hours = 2*60 = 120 minutes
0.1523 = 2.303/120[log (a/a-x)]
log (a/a-x)= (0.1523*120)/2.303
= 7.94
a/a-x = Antilog(7.94)
a/a-x=87096358
Let initial concentration = 100
100/100-x = 87096358
1.14*10-6 = 100-x
x = 100-(1.14*10-6)
x= 99.99
Thus 99.99% of reactant is consumed in reaction and (100-99.99) = 0.01% of reactant left in the reaction.
Order of a reaction is the power of concentration terms for those changes the rate of a reaction.
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