A reaction was run with two different initial concentrations of reactants A and B: Experiment A / M B...

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Chemistry

A reaction was run with two different initial concentrations ofreactants A and B:
Experiment
A / M
B / M
rate_B / (M/sec)
1
0.00022
0.00046
0.0841
2
0.00022
0.00166
0.3035
What is the order of the reaction with respect to B?
A reaction was run with two different initial concentrations ofreactants A and B:
Experiment
A / M
B / M
rate_B / (M/sec)
1
0.00057
0.00067
0.0000825
2
0.00057
0.00621
0.0657
What is the order of the reaction with respect to B?

Answer & Explanation Solved by verified expert

4.5 Ratings (962 Votes)

let r= k*[A]^x [B]^y
put the values from both experiments to get 2 equations
0.0841 = k* (0.00022)^x * (0.00046)^yÂ Â Â Â Â Â Â Â Â Â .....eqn 1
0.3035 = k* (0.00022)^x * (0.00166)^yÂ Â Â Â Â Â Â Â Â Â .....eqn 2

divide eqn 2 by 1

0.3035/0.0841 = (0.00166/0.00046)^y
3.61 = (3.61)^y
so,y=1

the order of the reaction with respect to B =1

let r= k*[A]^x [B]^y
put the values from both experiments to get 2 equations
0.0000825 = k* (0.00057)^x * (0.00067)^yÂ Â Â Â Â Â Â Â Â Â .....eqn 1
0.0657= k* (0.00057)^x * (0.00621)^yÂ Â Â Â Â Â Â Â Â Â .....eqn 2

divide eqn 2 by 1

0.0657/0.0000825 = (0.00621/0.00067)^y
796.34 = (9.27)^y
take log on both sides
log (796.34) = y * log (9.27)
y= log (796.34) / log(9.27)
=3
so,y=3

the order of the reaction with respect to B =3

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Experiment	A / M	B / M	rate_B / (M/sec)
1	0.00022	0.00046	0.0841
2	0.00022	0.00166	0.3035