The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. 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 1) Determine the order of the reaction. (6 pts.)
2) Find its rate constant. 19 pts.) Note: no unit is needed, just the numerical answer. Hint: convert your minutes to seconds.

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

its first and third is your answer i did the qustion the fourth isnt right