Answer:
True
Explanation:
Ethers react with HI to form the corresponding alcohols and alkyl iodides.
Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes reaction with HI to form ethyl iodide.
<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>
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