the following data were collected at 20C for thereaction of bromphenol blue (HBP
ID: 770201 • Letter: T
Question
the following data were collected at 20C for thereaction of bromphenol blue (HBPB2-) and hydroxide ions(OH-), shown in equation below HBPB2-+OH- BPB3-+H2O the times required to consume a small butconstant amount of HBPB2- at varying inititialHBPB2- and OH-concentrations were measuredand are recorded below. determination time,s 1 (7.22*10-6) (1.00) 75 2 (7.22*10-6) (0.25) 290 3 (3.63*10-6) (1.00) 152 A) DETERMINE THE REACTION ORDER WITH RESPECT TOOH-. B) DETERMINE THE REACTION ORDER WITH RESPECTTO HBPB2-. C) WRITE THE RATE LAW FOR THE REACTION OFHBPB2- WITH OH-Explanation / Answer
I can't see your table to properly answer your question. If you could post it with more clear formatting, I could give you a definite answer.
Alternitively I could just tell you how to do it.
First, look at what happened to the time, when you doubled the concentration of OH -. What happened? Was it cut in half, was it 1/4 of the previous time? Did it stay the same. If it stayed the same, you will give [OH] and exponent of 0 in the rate law. If it was cut in half, you will give [OH] an exponent of 1. If it was cut to 1/4 the previous time, you will give [OH] an exponent of 2.
Next, you do the exact same thing with HBPB2-.
Finally, you will write the rate law.
It will be in the form r = k[OH]^(a)[HBPB2-]^(b)
By now, you should know what to plug in for a and b.
You also, have data using different concentrationsof OH- and HBPB2-, so you can plug in those concentrations and the resulting rate and solve for k.
Good Luck.
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