We’re being asked to **determine the rate constant (k)** of a second-order reaction:

A → B + C

The concentration of A decreases by 80% (from 100% - 20% = 80%) starting 0.1 M in 26.2 minutes. 80% of 0.1 M is:

0.1 M x 0.8 = 0.08 M

The ** integrated rate law** for a second-order reaction is as follows:

$\overline{)\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{t}}}{\mathbf{=}}{\mathbf{kt}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}}$

where:

**[A] _{t}** = concentration at time t

**k** = rate constant

**t** = time

**[A] _{0}** = initial concentration

The reaction A => B + C is second order in A. When the initial [A] = 0.100 M, the reaction is 20.0% complete in 26.2 minutes. Calculate the value of the rate constant (in L/min·mol).

a. 8.22x10^{-2 }

b. 9.54x10^{-2}

c. 1.53

d. 1.05

e. 0.91

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