Question

Two chemicals \(c_{1}\) and \(c_{2}\) react to form a third chemical \(c_{3} .\) The rate of change of the number of pounds of \(c_{3}\) formed is proportional to the amounts of \(c_{1}\) and \(c_{2}\) present at any instant. The formation of \(c_{3}\) requires \(3 \mathrm{lb}\) of \(c_{2}\) for each pound of \(c_{1} .\) Suppose initially there are \(10 \mathrm{lb}\) of \(c_{1}\) and \(15 \mathrm{lb}\) of \(c_{2}\) present, and that \(5 \mathrm{lb}\) of \(c_{3}\) are formed in 15 minutes. (a) Find the amount of \(c_{3}\) present at any time. (b) How many \(\mathrm{lb}\) of \(c_{3}\) are present after 1 hour? Suggestion Let \(x\) be the number of pounds of \(c_{3}\) formed in time \(t>0 .\) The formation requires three times as many pounds of \(c_{2}\) as it does of \(c_{1}\), so to form \(x\) lb of \(c_{3}, 3 x / 4\) lb of \(c_{2}\) and \(x / 4 \mathrm{lb}\) of \(c_{1}\) are required. So, from the given initial amounts, there are \(10-x / 4 \mathrm{lb}\) of \(c_{1}\) and \(15-3 x / 4 \mathrm{lb}\) of \(c_{2}\) present at time \(t\) when \(x\) lb of \(c_{9}\) are formed. Thus we have the differential equation $$ \frac{d x}{d t}=k\left(10-\frac{x}{4}\right)\left(15-\frac{3 x}{4}\right) $$ where \(k\) is the constant of proportionality. We have the initial condition $$ x(0)=0 $$ and the additional condition $$ x(15)=5. $$

Short answer

To find the amount of chemical $c_3$ at any time $t$, we solved the differential equation and found: \[ x(t) = 40 - 4 \left(10 - \frac{-4(kt + C)}{(2(10-\frac{x}{4})(15 - \frac{3x}{4}))}\right) \] After 1 hour, the amount of chemical $c_3$ present is: $x(1) = 40 - 4 \left(10 - \frac{-4(k + C)}{(2(10-\frac{x}{4})(15 - \frac{3x}{4}))} \right)\bigg|_{t=1}$.

Step-by-step solution

Set up the differential equation

The given differential equation is: \[ \frac{dx}{dt} = k\left(10 - \frac{x}{4}\right)\left(15 - \frac{3x}{4}\right) \] with initial condition \(x(0) = 0\), where \(x\) is the amount of chemical \(c_3\) formed in time \(t\).

Identify additional condition

The problem states that after 15 minutes (which is equal to 0.25 hours), 5 lb of \(c_3\) are formed. This translates to an additional condition for the differential equation: \[ x(0.25) = 5 \]

Solve the differential equation

To solve the differential equation, we'll first separate variables: \[ \frac{dx}{(10 - x/4)(15 - 3x/4)} = k \, dt \] Now, integrate both sides with respect to their respective variables: \[ \int \frac{dx}{(10 - x/4)(15 - 3x/4)} = \int k \, dt \] To make the integration on the left-hand side easier, we perform a substitution, let \(u = 10 - x/4\), then \(x = 40 - 4u\) and \(dx = -4 \, du\). Then we can rewrite the integral as: \[ \int \frac{-4 \, du}{u(30 - 2u)} = \int k \, dt \] Now integrate both sides: \[ -4 \int \frac{du}{u(30 - 2u)} = kt + C \] Now we need to find the constant of integration \(C\). Use the initial condition, \(x(0) = 0\): \[ -4 \int \frac{du}{u(30 - 2u)} \Bigg|_{x = 0} = kt + C \Bigg|_{t = 0} \] This gives \[ C = -4 \int \frac{du}{u(30 - 2u)} \Bigg|_{u = 10} \] Now we need to find the constant of proportionality, \(k\), using the additional condition, \(x(0.25) = 5\): \[ -4 \int \frac{du}{u(30 - 2u)} \Bigg|_{u = 7.5} = kt + C \Bigg|_{t = 0.25} \] Solving for \(k\), we get: \[ k = \frac{-4 \int \frac{du}{u(30 - 2u)} \Bigg|_{u = 7.5} - C}{0.25} \]

Rewrite the integral in terms of x

We have integrated the equation in terms of \(x\), so we'll now rewrite the equation in terms of \(x\) by substituting \(u\) with \(10-x/4\): \[ -4 \int \frac{du}{u(30 - 2u)} = kt + C \Bigg|_{u = 10 - x/4} \] Now solve for \(x\) to find the amount of \(c_3\) present at any time \(t\): \[ x = 40 - 4u \Bigg|_{ u = 10 - \frac{-4(kt + C)}{(2(u)(30 - 2u))} } \]

Evaluate the amount of \(c_3\) after 1 hour

Set \(t = 1\) (hour) in the expression for \(x\) obtained in the previous step: \[ x(1) = 40 - 4u \Bigg|_{ u = 10 - \frac{-4(k + C)}{(2(u)(30 - 2u))} }\bigg|_{t=1} \] By evaluating this expression, we obtain the amount of chemical \(c_3\) present after 1 hour. This concludes the exercise.

Key concepts

Chemical Reaction Rate

Understanding the rate of a chemical reaction is crucial in various fields such as chemical engineering, environmental science, and pharmacology. In our exercise, we examine a reaction where two chemicals, denoted as \(c_{1}\) and \(c_{2}\), combine to form a third chemical \(c_{3}\). The rate at which \(c_{3}\) is formed is proportional to the instantaneous quantities of \(c_{1}\) and \(c_{2}\).

This proportionality is encapsulated in a differential equation that correlates the rate of formation, \(\frac{dx}{dt}\), with the amounts of \(c_{1}\) and \(c_{2}\) still available to react. Since the reaction requires a specific ratio of \(c_{1}\) to \(c_{2}\) (1 lb to 3 lb in this context), the differential equation will reflect this stoichiometry. Understanding the equation allows us to predict the behavior of the reaction over time and determine the quantity of \(c_{3}\) produced at any given moment.

Separable Differential Equations

A separable differential equation is one in which the variables can be separated on either side of the equation, allowing for integration. In the case of our chemical reaction, we are dealing with the equation:
\[\frac{dx}{dt} = k\left(10 - \frac{x}{4}\right)\left(15 - \frac{3x}{4}\right)\]Where \(k\) is the proportionality constant, \(x\) represents the amount of product formed, and \(t\) is time. As illustrated by the steps in the solution, the first step towards solving this equation is to separate the variables \(x\) and \(t\), placing them on opposite sides of the equation.

Next, we integrate both sides to find a solution. The integration process might require substitution to simplify the terms, which is a common technique when dealing with more complex separable equations. Finally, we solve for the constant of integration using the given initial conditions. This process unveils the relationship between the reactant quantities and time, allowing us to track the formation of \(c_{3}\) throughout the reaction.

Initial Value Problems

Initial value problems are a category of differential equation problems where the solution must satisfy a given initial condition. In our exercise, we are provided with the initial condition that \(x(0) = 0\), indicating that at the start of the reaction (\(t = 0\)), no product \(c_{3}\) has been formed. This information is essential as it serves as the starting point for our integration.

Furthermore, the problem statement includes another condition: \(x(15) = 5\). This offers a snapshot of the system 15 minutes into the reaction, where 5 lb of \(c_{3}\) have been produced. It serves as a valuable checkpoint to determine the proportionality constant \(k\), which is a critical parameter of the reaction rate. In essence, initial value problems provide specific details that allow for the full characterization of the system's behavior over time, making them particularly useful in predicting the outcomes of real-world scenarios, such as chemical reactions.