It's difficult to write it down, but I'll attach you a good example of hydroboration of indene. I hope you'll find it helpful.
I don't know if I'm doing this right can someone help me out
1 answer:
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Answer:
1) The order of the reaction is of FIRST ORDER
2) Rate constant k = 5.667 × 10 ⁻⁴
Explanation:
From the given information:
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.
liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.
The following data was obtained:
t/min 0 10 20 30 40 ∞
conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312
For a first order reaction:
where :
K = proportionality constant or the rate constant for the specific reaction rate
t = time of reaction
= initial concentration at time t
= final concentration at time t
= concentration at time t
To start with the value of t when t = 10 mins
When t = 20
When t = 30
When t = 40
We can see that at the different time rates, the rate constant of all have similar constant values
As such :
Rate constant k = 0.034 min⁻¹
Converting it to seconds ; we have :
60 seconds = 1 min
∴
0.034 min⁻¹ =(0.034/60) seconds
= 5.667 × 10 ⁻⁴ seconds
Rate constant k = 5.667 × 10 ⁻⁴