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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 Answer

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Answer:

1) The order of the reaction is of FIRST ORDER

2) Rate constant k = 5.667 × 10 ⁻⁴

Step-by-step 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 = (1)/(t) \ In ( (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 = (1)/(10) \ In ( (0.312 - 0)/(0.312 - 0.089))


K_1 = (1)/(10) \ In ( (0.312 )/(0.223))


K_1 =0.03358 \ min^(-1)


K_1 \simeq 0.034 \ min^(-1)

When t = 20


K_2= (1)/(20) \ In ( (0.312 - 0)/(0.312 - 0.153))


K_2= 0.05 * \ In ( 1.9623)


K_2=0.03371 \ min^(-1)


K_2 \simeq 0.034 \ min^(-1)

When t = 30


K_3= (1)/(30) \ In ( (0.312 - 0)/(0.312 - 0.200))


K_3= 0.0333 * \ In ( (0.312)/(0.112))


K_3= 0.0333 * \ 1.0245


K_3 = 0.03412 \ min^(-1)


K_3 = 0.034 \ min^(-1)

When t = 40


K_4= (1)/(40) \ In ( (0.312 - 0)/(0.312 - 0.230))


K_4=0.025 * \ In ( (0.312)/(0.082))


K_4=0.025 * \ 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

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