What is the numerical value of the rate constant for a reaction in which [A] is reduced from 0.380 M to 0.215 M in a span of 26.3 s? Assume the units of the rate constant are s-1. Please enter your response to three of significant figures using E notation. For example, to report 1347.54 to three significant figures, type: 1.35E3

k = 2.17E-2 s-1

Explanation:

Initial Concentration [A]o = 0.380 M

Final Concentration [A] = 0.215 M

time, t = 26.3 s

Unit of rate constant =  s-1

Rate constant = ?

The units of the rate constant depends on the order of the reaction. This is a first order reaction due to the units.

The integral rate law for a first order reaction is;

ln[A] = ln[A]o - kt

ln(0.215) = ln(0.380) - k(26.3)

ln(0.215) - ln(0.380) = - k(26.3)

- k(26.3) = -0.57

k = 0.57 / 26.3 = 0.02167

Reporting to three significant figures using E notation;

k = 2.17E-2 s-1

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g