Asked by pancake
1. The overall order for the reaction, A+B→C, is 2. A student tests this by measuring the reaction rate at one concentration of A and B, then doubling both concentrations at the same time and re-measuring the rate, which quadrupled. The student concludes that the data support the hypothesis that the above rate law is rate=k[A][B]. Analyze this.
a. The data do support the hypothesis but each trial should be repeated.
b. The data contradict the hypothesis because the rate should have doubled, not quadrupled. ***
c. The data support the hypothesis but a trial holding one reactant constant is needed.
d. The data contradict the hypothesis but testing by changing a single reactant will verify the hypothesis.
2. Determine the overall reaction order for the reaction, A+B→2C using the table of data below:
Experiment Initial [A]/M Initial [B]/M Initial rate/M s-1
1 0.010 0.040 0.0031
2 0.010 0.020 0.0015
3 0.020 0.020 0.0032
a. 0
b. 1
c. 2
d. 3 ***
a. The data do support the hypothesis but each trial should be repeated.
b. The data contradict the hypothesis because the rate should have doubled, not quadrupled. ***
c. The data support the hypothesis but a trial holding one reactant constant is needed.
d. The data contradict the hypothesis but testing by changing a single reactant will verify the hypothesis.
2. Determine the overall reaction order for the reaction, A+B→2C using the table of data below:
Experiment Initial [A]/M Initial [B]/M Initial rate/M s-1
1 0.010 0.040 0.0031
2 0.010 0.020 0.0015
3 0.020 0.020 0.0032
a. 0
b. 1
c. 2
d. 3 ***
Answers
Answered by
pancake
2. Here's a better version of the table.
1 | 0.010 | 0.040 | 0.0031
2 | 0.010 | 0.020 | 0.0015
3 | 0.020 | 0.020 | 0.0032
1 | 0.010 | 0.040 | 0.0031
2 | 0.010 | 0.020 | 0.0015
3 | 0.020 | 0.020 | 0.0032
Answered by
DrBob222
I don't believe either answer is correct. What is your rationale for each?
Answered by
pancake
1. since the order of reaction was 2, I thought it would double.
2. I change my answer to option C. Is that correct?
T 1 & 2: 0.0031 / 0.0015 = k (0.010)^m (0.040)^n / k (0.010)^m (0.020)^n. N = 1
T 2 & 3: 0.0015 / 0.0032 = (0.010/ 0.020)^n. M= 1
m + n = 1 + 1 = 2
2. I change my answer to option C. Is that correct?
T 1 & 2: 0.0031 / 0.0015 = k (0.010)^m (0.040)^n / k (0.010)^m (0.020)^n. N = 1
T 2 & 3: 0.0015 / 0.0032 = (0.010/ 0.020)^n. M= 1
m + n = 1 + 1 = 2
Answered by
DrBob222
Yes, 2c is correct.
On question 1, if you double one reactant that will double the rate. True. But if you double BOTH that will double BOTH and 2 x 2 = 4. So could 1b be correct?
On question 1, if you double one reactant that will double the rate. True. But if you double BOTH that will double BOTH and 2 x 2 = 4. So could 1b be correct?
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