(a) What is the rate law for this reaction? 1) Rate = k [ClO 2 (aq)] [OH - (aq)]

Question

(a) What is the rate law for this reaction?

1) Rate = k [ClO2(aq)] [OH-(aq)]

2) Rate = k [ClO2(aq)]2 [OH-(aq)]    

3) Rate = k [ClO2(aq)] [OH-(aq)]2

4) Rate = k [ClO2(aq)]2 [OH-(aq)]2

5) Rate = k [ClO2(aq)] [OH-(aq)]3

6) Rate = k [ClO2(aq)]4 [OH-(aq)]

(b) What is the value of the rate constant?

____________________.


(c) What is the reaction rate when the concentration of ClO2(aq) is 0.0567 M and that of OH-(aq) is 0.0760 M if the temperature is the same as that used to obtain the data shown above?

______________M/s

Experiment [ClO2(aq)] (M) [OH-(aq)] (M) Rate (M/s) 1 0.0302 0.0302 0.0112 2 0.0302 0.0604 0.0224 3 0.0604 0.0302 0.0448 4 0.0604 0.0604 0.0897

Explanation / Answer

a)
see experiment 1 and 2:
[ClO2] is constant and [OH-] doubles, rate also doubles
So, order of OH- is 1

see experiment 1 and 3:
[ClO2] doubles and [OH-] is constant, rate becomes 4 times
So, order of ClO2 is 2

rate law is:
Rate = k [ClO2(aq)]2 [OH-(aq)]   
Answer: 2)

b)
Rate = k [ClO2(aq)]2 [OH-(aq)]   
put values from experiment 1
0.0112= k (0.0302)^2 * 0.0302
k = 406.6 M-2s-1

c)
Rate = k [ClO2(aq)]2 [OH-(aq)]   
= 406.6 * (0.0567)^2 * (0.0760)
= 0.099 M/s

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