Question

For a first-order reaction \(a \mathrm{~A} \longrightarrow\) products, where \(a \neq 1\), the rate is \(-\Delta[\mathrm{A}] / a \Delta t,\) or in derivative notation, \(-\frac{1}{a} \frac{d[\mathrm{~A}]}{d t}\) Derive the integrated rate law for the first-order decomposition of \(a\) moles of reactant.

Short answer

Question: Derive the integrated rate law for a first-order reaction involving the decomposition of "a" moles of reactant A, where "a" is not equal to 1. The rate is given as \(-\frac{1}{a} \frac{d[\mathrm{A}]}{d t}\). Answer: The integrated rate law for the first-order decomposition involving "a" moles of reactant A is given by: \(\ln{\left(\frac{[\mathrm{A}]}{[\mathrm{A}]_0}\right)} = \frac{1}{a}([\mathrm{A}]_0 - [\mathrm{A}])\)

Step-by-step solution

Rewrite the rate equation in terms of concentrations and time

To derive the integrated rate law, we'll need to rewrite the given rate equation in terms of concentrations of A and time. The rate equation is: \(-\frac{1}{a} \frac{d[\mathrm{A}]}{d t}\)

Separate variables

In order to integrate both sides, we will need to separate the variables. This can be done by multiplying both sides with "a" and taking the concentration of A to the left side and time "dt" to the right side. \(-d[\mathrm{A}] = a\frac{d[\mathrm{A}]}{[A]} dt\)

Integrate both sides

Now that the variables are separated, we can integrate both sides. When integrating, it's important to include the proper bounds. We'll use the initial concentration of A as \([\mathrm{A}]_0\) at time \(t = t_0\) and the final concentration of A as \([\mathrm{A}]\) at time \(t = t\). The integration would be: \(\int_{[\mathrm{A}]_0}^{[\mathrm{A}]} -d[\mathrm{A}] = \int_{t_0}^{t} a\frac{d[\mathrm{A}]}{[A]} dt\)

Evaluate the integrals

Now we'll evaluate both integrals. On the left side, the integral will result in \([\mathrm{A}]_0 - [\mathrm{A}]\). On the right side, the integral will result in \(a \ln{[\mathrm{A}]}\). So we have: \([\mathrm{A}]_0 - [\mathrm{A}] = a \ln{[\mathrm{A}]} - a \ln{[\mathrm{A}]_0}\)

Isolate \([\mathrm{A}]\)

Now we need to isolate \([\mathrm{A}]\) on the left side. First, move \([\mathrm{A}]\) to the right side and then divide both sides by \(-a\): \([\mathrm{A}]_0 = [\mathrm{A}] - a (\ln{[\mathrm{A}]} - \ln{[\mathrm{A}]_0})\) \([\mathrm{A}] = [\mathrm{A}]_0 + a (\ln{[\mathrm{A}]} - \ln{[\mathrm{A}]_0})\) Now use the property of logarithms, \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\), to rewrite the logarithmic part: \([\mathrm{A}] = [\mathrm{A}]_0 + a \ln{\left(\frac{[\mathrm{A}]}{[\mathrm{A}]_0}\right)}\) Finally, isolate the term \(\ln{\left(\frac{[\mathrm{A}]}{[\mathrm{A}]_0}\right)}\) and divide both sides by "a": \(\ln{\left(\frac{[\mathrm{A}]}{[\mathrm{A}]_0}\right)} = \frac{1}{a}([\mathrm{A}]_0 - [\mathrm{A}])\) The resulting equation represents the integrated rate law for the first-order decomposition involving "a" moles of reactant A.