Question

The rate of decrease of a reactant is proportional to the square of the amount present. During a particular reaction, \(40 \%\) of the initial amount of this chemical remained after \(10 \mathrm{sec}\). How long will it take before only \(25 \%\) of the initial amount remains?

Short answer

Answer: It takes 40 seconds for the chemical reactant to reach 25% of its initial amount.

Step-by-step solution

Set up the differential equation

The problem states that the rate of decrease of the reactant is proportional to the square of the amount present. Let \(y(t)\) be the amount of reactant at time \(t\), and \(k\) be the constant of proportionality. The rate of decrease can be represented by the derivative \(-\frac{dy}{dt}\), and as a result, you can form the following differential equation: \[ -\frac{dy}{dt} = ky^2 \]

Rearrange and integrate

Separate the variables \(y\) and \(t\) on the two sides of the equation, and then integrate both sides: \[ -\int\frac{1}{y^2} dy = k\int dt \] After performing the integration, you will get: \[ \frac{1}{y} = kt + C \] Where \(C\) is the constant of integration.

Use given information to find constants

At \(t = 0\), we assume \(y(0) = y_0\) to be the initial amount of reactant, and when \(t = 10\), according to the problem, \(y(10) = 0.4y_0\). Plugging these values into the equation, we can find \(k\) and \(C\). For \(t = 0\), \[ \frac{1}{y_0} = k(0) + C \Rightarrow C = \frac{1}{y_0} \] For \(t = 10\), \[ \frac{1}{0.4y_0} = k(10) + \frac{1}{y_0} \] Solving for \(k\), we get: \[ k = \frac{1}{30 y_0} \]

Set up the equation to find the desired time

Now that we have \(k\) and \(C\), we can set up the equation to find the time when only \(25\%\) of the initial amount remains, i.e., \(y(t) = 0.25y_0\): \[ \frac{1}{0.25y_0} = \frac{1}{30 y_0}t + \frac{1}{y_0} \]

Solve for the desired time

Now, let's solve for \(t\) in the following equation: \[ \frac{1}{0.25y_0} = \frac{1}{30 y_0}t + \frac{1}{y_0} \] Subtract \(\frac{1}{y_0}\) from both sides and multiply both sides by \(30y_0\) to eliminate the fractions: \[ 30y_0\left(\frac{1}{0.25y_0} - \frac{1}{y_0}\right) = t \] Simplifying the equation, you'll find the desired time: \[ t = \frac{30y_0}{0.25y_0 - y_0} = -\frac{30}{0.75} = 40\,\mathrm{sec} \] So it will take \(\boxed{40\,\mathrm{sec}}\) for only \(25\%\) of the initial amount of reactant to remain.

Key concepts

Rate of Change

The concept of Rate of Change is vital in understanding how quantities vary over time. For example, in our exercise, the rate at which a reactant decreases is linked to a certain variable related to its concentration. The rate of change often involves derivatives, symbolized as \(\frac{dy}{dt}\). This derivative measures how the amount \(y(t)\), of the reactant, changes with respect to time \(t\).

In certain chemical reactions, such as the one in our exercise, this rate is linked to how much reactant remains. Here, the rate of decrease is proportional to the square of the amount present. This intrinsic relationship is captured by the differential equation formed. Understanding rate of change helps in predicting future behavior of the reactant given its current state.

Separable Equations

Separable Equations are a type of differential equation that can be rearranged to separate variables on opposite sides. This is crucial for solving many differential equations. In our problem, an equation was derived, \(-\frac{dy}{dt} = ky^2\), which was later rearranged by separating \(y\) and \(t\) to opposite sides.

To solve this, you perform integration on both sides. The integration transforms the differential equation into a form where direct substitution of initial conditions is possible. By separating variables, tackling complex questions becomes manageable, allowing for systematic integration to find solutions.

Initial Value Problems

Initial Value Problems are differential equations which must satisfy specific conditions at initial points. These problems describe dynamical systems and if given initial conditions, like the amount of reactant at \(t = 0\), they help determine unique solutions.

In the exercise, knowing that \(40\%\) of the reactant remains after \(10\) seconds helps determine constants, resulting in a precise solution. Using initial conditions is essential for modeling behaviors precisely and for making accurate predictions regarding future states of reaction processes.

Proportional Relationships

Proportional Relationships occur when two quantities maintain a constant ratio to each other. In mathematics and science, this frequently appears, as in our problem, where the rate of decrease of a reactant is directly proportional to the square of the reactant present. This is expressed using a constant \(k\), altering how the proportionality shows in equations.

These relationships underpin many natural phenomena and help in forming differential equations that describe how quantities evolve. Recognizing and exploiting these relationships allows for an understanding of complex systems through simpler mathematical expressions.