A reaction was run with two different initial concentrations of reactants A and

Question

A reaction was run with two different initial concentrations of reactants A and B:

Experiment

A / M

B / M

rateB / (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 of reactants A and B:

Experiment

A / M

B / M

rateB / (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?

Experiment

A / M

B / M

rateB / (M/sec)

1

0.00022

0.00046

0.0841

2

0.00022

0.00166

0.3035

Explanation / Answer

1)

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)^y           .....eqn 1
0.3035 = k* (0.00022)^x * (0.00166)^y           .....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

2)

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)^y           .....eqn 1
0.0657= k* (0.00057)^x * (0.00621)^y           .....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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