{Use of Tech} Chemical rate equations Let y(t) be t he concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation dy/dt=-ky^n for t≥0, where k>0 is a rate constant and the positive integer n is the order of the reaction.
b. Solve the initial value problem for a second-order reaction (n=2) assuming y(0)=y0.

Verified step by step guidance

1. Identify the given differential equation for the second-order reaction: \(\frac{dy}{dt} = -k y^{2}\), where \(k > 0\) and the initial condition is \(y(0) = y_0\).
2. Rewrite the differential equation to separate variables, moving all terms involving \(y\) to one side and \(t\) to the other: \(\frac{dy}{y^{2}} = -k \, dt\).
3. Integrate both sides: integrate \(\int \frac{dy}{y^{2}}\) with respect to \(y\) and \(\int -k \, dt\) with respect to \(t\).
4. After integration, apply the initial condition \(y(0) = y_0\) to solve for the constant of integration.
5. Finally, solve the resulting equation for \(y\) explicitly as a function of \(t\) to express the concentration \(y(t)\).

Key Concepts

Separable Differential Equations

A separable differential equation can be written as dy/dt = g(t)h(y), allowing variables y and t to be separated on opposite sides. This enables integration with respect to each variable independently, which is essential for solving equations like dy/dt = -ky^n.

Initial Value Problem (IVP)

An initial value problem specifies the value of the unknown function at a particular point, such as y(0) = y0. This condition allows us to find the particular solution to a differential equation that fits the given starting concentration.

Second-Order Reaction Rate Law

For a second-order reaction (n=2), the rate equation is dy/dt = -ky^2. Solving this involves integrating the reciprocal of y squared, leading to a solution that describes how concentration decreases over time, typically resulting in a rational function of t.