# Problem: The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is(b) second order with respect to A?

###### FREE Expert Solution

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 (unknown)

[A]0 = initial concentration

Calculate k:

The half-life of a second-order reaction is given by:

$\overline{){{\mathbf{t}}}_{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{k}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}}$

where:

k = rate constant

[A]0 = initial concentration

${\mathbf{t}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{k}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}$

k = 0.7843 L/mol • min

Solving for time:

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

95% (407 ratings) ###### Problem Details

The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is

(b) second order with respect to A?

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