5. The rate constant for the first-order isomerization of cyclopropane to propen

Question

5. The rate constant for the first-order isomerization of cyclopropane to propene is 0.024 min1 at 760K. (5 points each) Cyclopropane (C,H)Propene (C,H) 2 Calculate the half life of cyclopropane at 760 K. Given an initial cyclopropane concentration of 0.20 moles/L, calculate the time it will take for the cyclopropane concentration to decrease to 0.075 moles/L a. b. c. Given an initial cyclopropane concentration of 0.16 moles/L, calculate the concentration of d) How long does it take for the cyclopropane concentration to decrease to 20% of its initial e.) You start a reaction with 11.0 grams of cyclopropane in a 2.5 L flask. How many grams of cyclopropane remaining after 1.5 hours. (Watch the units of k vs t. They have to be the same to cancel! Likewise the units of A and Ao must be the same.) value? propene are produced in the first hour (of reaction)?

Explanation / Answer

Ans 5

The reaction is of first order

Rate constant k = 0.024 min-1

Part a

Half life for first order reaction

t1/2 = 0.6932/k

= 0.6932/0.024

= 28.883 min

Part b

Initial concentration [A0] = 0.20 M

Final concentration [A] = 0.075 M

Time t =?

For first order reaction

ln ([A] / [A0]) = - k x t

ln (0.075/0.20) = - 0.024 x t

t = 40.87 min

Part c

Initial concentration [A0] = 0.16 M

Time t = 1.5 hr x 60min/hr = 90 min

Final concentration [A] =?

For first order reaction

ln ([A] / [A0]) = - k x t

[A] = [A0] exp ( - k x t)

= (0.16 M) x exp (-0.024 min-1 x 90 min)

= 0.0185 M

Part d

Let Initial concentration [A0] = 100 M

Final concentration [A] = 0.20 x 100 = 20 M

Time t =?

ln ([A] / [A0]) = - k x t

ln (20/100) = - 0.024 x t

t = 67.10 min

Part e

Moles of cyclopropane (C3H6) = mass/molecular weight

= 11g / 42.08 g/mol

= 0.2614 mol

Initial concentration [A0] = moles/Volume

= 0.2614 mol / 2.5 L

= 0.10456 M

Final concentration [A] =?

Time t = 1 hr = 60 min

For first order reaction

ln ([A] / [A0]) = - k x t

[A] = [A0] exp ( - k x t)

= (0.10456 M) x exp (-0.024 min-1 x 60 min)

= 0.02477 M

Final moles of propene = molarity x volume

= 0.02477 mol/L x 2.5 L

= 0.06193 mol

Mass of propene produced = moles x molecular weight

= 0.06193 mol x 42.08 g/mol

= 2.61 g

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