1) A reaction was run with two different initial concentrations of reactants A a
ID: 889459 • Letter: 1
Question
1)
A reaction was run with two different initial concentrations of reactants A and B:
Experiment
A / M
B / M
rateB / (M/sec)
1
0.00082
0.00053
0.0737
2
0.00082
0.00228
0.3169
What is the order of the reaction with respect to B?
2)
A reaction was run with two different initial concentrations of reactants A and B:
Experiment
A / M
B / M
rateB / (M/sec)
1
0.00059
0.00064
0.000865
2
0.00059
0.00697
0.1026
What is the order of the reaction with respect to B?
Experiment
A / M
B / M
rateB / (M/sec)
1
0.00082
0.00053
0.0737
2
0.00082
0.00228
0.3169
Explanation / Answer
1) rate equation for the reaction is rate = k [B]n, where n = order with respect to B
0.0737 = k 0.00053n ---------(1)
0.3169 = k 0.00228n --------(2) , Now divide equation (1) by (2)
0.0737/0.3169 = (0.00053/0.00228)n
0.2325 = (0.2325)n
n = 1 , so first order reaction with respect to B
= k 0.00053 ---------(1)
0.3169 = k 0.00228 --------(2)
0.0737/0.3169 = (0.00053/0.00228)n
0.2325 = (0.2325)n
n = 1 , so first order reaction with respect to B
2) rate equation for the reaction is rate = k [B]n
0.000865 = k 0.00064n ---------(1)
0.1026 = k 0.00697n --------(2) , Now divide equation (1) by (2)
0.000865/0.1026 = (0.00064/0.00697)n
0.00843 = (0.09182)n
(0.09182)1/2 = (0.09182)n
n = 1/2 , so order of the reaction with respect to B is 1/2
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