I need help with these questions please, and with the calculations for the table
ID: 958007 • Letter: I
Question
I need help with these questions please, and with the calculations for the table
Concentrations of Reactants- obtain the concentrations from the bottles located within the Oxalic Acid]: _oasso Data Table 1 Experiment Temperature Average Reaction Tie (se Reaction Time (sec) Trial 2 ads Trial 3 Trial 1 Data Analysis and Calculations Answer the following questions in order to d the question involves any calculations, be sure to include them when answering the questions etermine the rate law and rate constant for this reaction, 1. To determine the reaction order for each species, it is important to compare how the rate of the reaction changed as the concentration of one of the reactants was changed, and the other reactants were kept constant. Which two experiments can be used to determine the order for oxalic acid? Justify your answe a. b. Which two experiments can be used to determine the order for MnO4? Justify your answer. Before the rates of each reaction can be determined, the change in concentration of each reactant must be determined. 2. To determine the initial concentration of each reactant, a dilution calculation must be performed in order to account for the concentration change of each reactant as othe solutions are added. Fill in Data Table 2 below with the initial concentrations of oxali acid and permanganate. Be sure to show any calculations required. a.Explanation / Answer
Solution:
[Oxalic acid] = 0.7550 M
[KMnO4] = 0.1300 M
1. Rate = reactant / t
For Oxalic Acid
Experiment
Rate (Trial 1)
Rate (Trial 2)
Rate (Trial 3)
1
= 0.7550 / (266 – 0)
=0.002833 M/sec
=0.7550/(274-0)
=0.00275 M/sec
=0.7550/(272-0)
=0.002775 M/sec
2
=0.7550/(169-0)
=0.004467 M/sec
=0.7550/((168-0)
=0.004494 M/sec
=0.7550/(176-0)
=0.00428 M/sec
3
=0.7550/(247-0)
=0.003056 M/sec
=0.7550/(244-0)
=0.003094 M/sec
=0.7550/(241-0)
=0.003132 M/sec
The experiment 2 and 3 is favoured as it clear from the above table rate of reaction is high for these experiments while comparing with 1.
For KMnO4
Experiment
Rate (Trial 1)
Rate (Trial 2)
Rate (Trial 3)
1
= 0.1300 / (266 – 0)
=0.0004887 M/sec
=0.1300 /(274-0)
=0.0004744 M/sec
=0.1300 /(272-0)
=0.0004779 M/sec
2
=0.1300 /(169-0)
=0.0007692 M/sec
=0.1300 /((168-0)
=0.0007738 M/sec
=0.1300 /(176-0)
=0.0007386 M/sec
3
=0.1300 /(247-0)
=0.000526 M/sec
=0.1300 /(244-0)
=0.0005327 M/sec
=0.1300 /(241-0)
=0.0005394 M/sec
The experiment 2 and 3 is favoured as it clear from the above table rate of reaction is high for these experiments while comparing with 1.
2. Determination of the rate order
Determination of the rate order Oxalic acid
Experiment #
[Oxalic acid]initial
[Oxalic acid]final
[Oxalic acid]
Reaction time (sec)
t
Rate M/s
2
0
0.7550
0.1300
175
=0.7550/175
=0.004314
3
0
0.7550
0.1300
244
=0.7550/244
=0.003094
Determination of the rate order MnO4
Experiment #
[MnO4]initial
[MnO4]final
[ MnO4]
Reaction time (sec)
t
Rate M/s
2
0
0.1300
0.1300
175
0.0007428
3
0
0.1300
0.1300
244
0.0005327
3. Pre-lab data computation required for comparing the dataset.
Experiment
Rate (Trial 1)
Rate (Trial 2)
Rate (Trial 3)
1
= 0.7550 / (266 – 0)
=0.002833 M/sec
=0.7550/(274-0)
=0.00275 M/sec
=0.7550/(272-0)
=0.002775 M/sec
2
=0.7550/(169-0)
=0.004467 M/sec
=0.7550/((168-0)
=0.004494 M/sec
=0.7550/(176-0)
=0.00428 M/sec
3
=0.7550/(247-0)
=0.003056 M/sec
=0.7550/(244-0)
=0.003094 M/sec
=0.7550/(241-0)
=0.003132 M/sec