Question

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.

Answer 1

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:

$K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})$

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

$C_o$ = initial concentration at time t

$C _{\infty}$ = final concentration at time t

$C_t$ = concentration at time t

To start with the value of t when t = 10 mins

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})$

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})$

$K_1 =0.03358 \ min^{-1}$

$K_1 \simeq 0.034 \ min^{-1}$

When t = 20

$K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})$

$K_2= 0.05 \times \ In ( 1.9623)$

$K_2=0.03371 \ min^{-1}$

$K_2 \simeq 0.034 \ min^{-1}$

When t = 30

$K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})$

$K_3= 0.0333 \times \ In ( \dfrac{0.312}{0.112})$

$K_3= 0.0333 \times \ 1.0245$

$K_3 = 0.03412 \ min^{-1}$

$K_3 = 0.034 \ min^{-1}$

When t = 40

$K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})$

$K_4=0.025 \times \ In ( \dfrac{0.312}{0.082})$

$K_4=0.025 \times \ In ( 3.8048)$

$K_4=0.03340 \ min^{-1}$

We can see that at the different time rates, the rate constant of $k_1, k_2, k_3, and k_4$ 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