Question

First Order Chemical Reactions In some chemical reactions the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of \(\delta\)-glucono lactone into gluconic acid, for example, $$\frac{d y}{d t}=-0.6 y$$ when \(y\) is measured in grams and \(t\) is measured in hours. If there are 100 grams of a \(\delta\) -glucono lactone present when \(t=0,\) how many grams will be left after the first hour?

Short answer

After one hour, there will be approximately 54.88 grams of \(\delta\)-glucono lactone left.

Step-by-step solution

Solve the Diffential Equation

The given differential equation is in the form of \(\frac{d y}{d t} = k y\) where \(k = -0.6\). This is a simple first order homogeneous linear differential equation and can be solved by separation of variables, which gives: \(\int \frac{1}{y} dy = -0.6 \int dt\). Integrating both sides, we get \(\ln|y| = -0.6t + C\), where \(C\) is the constant of integration.

Find the Constant of Integration

We can find \(C\) by using the initial condition \(y(0) = 100\). Substituting these values into the equation we get \(\ln|100| = -0.6 * 0 + C\), which simplifies to \(C = \ln|100|\). Substituting \(C\) back into the equation, we get \(ln|y| = -0.6t + ln|100|\), which can be rewritten in exponential form as \(y = 100 e^{-0.6t}\), using that \(e^{ln|100|} = 100\). This is the function that models the amount of \(\delta\)-glucono lactone present at any given time.

Find the Quantity after One Hour

Substitute \(t = 1\) into the model to find the amount left after one hour: \(y(1) = 100 e^{-0.6 * 1} \approx 54.88 \, \text{grams}\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that describe the relationship between a function and its derivatives. They are a powerful tool used in science and engineering to model the behavior of dynamic systems, such as the rate of chemical reactions, population growth, or electrical circuits. A first-order differential equation involves the first derivative of a function and expresses how the rate of change of a function is related to the function itself.

In the context of chemical reactions, a differential equation can depict the rate at which a reactant is consumed or a product is formed. For instance, the equation \( \frac{d y}{d t} = -0.6 y \) represents a first-order reaction where the rate of change in the concentration (\( \frac{d y}{d t} \) is proportional to the current concentration (\(y\)). Solving this type of equation helps understand how the concentration of a reactant changes over time.

Exponential Decay

Exponential decay is a process by which a quantity decreases at a rate proportional to its current value. This type of behavior is ubiquitous in the natural world, occurring in scenarios like radioactive decay, cooling of hot objects, and, as in our chemical reaction example, the degradation of substances.

In the given problem, the decrease in the amount of \(\delta\)-glucono lactone over time is an example of exponential decay. The mathematical model for such decay is generally given by \(y = y_0e^{-kt}\), where \(y_0\) is the initial quantity, \(k\) is the decay constant, and \(t\) is time. In our case, \(y = 100e^{-0.6t}\) illustrates that the amount of \(\delta\)-glucono lactone decays exponentially as time progresses. This form allows us to calculate the remaining amount of the substance at any given time.

Separation of Variables

Separation of variables is a technique used to solve differential equations, where the variables are separated on different sides of the equation to facilitate integration. It's an essential tool to convert the differential equation into a more straightforward form that can be integrated to find a solution.

By moving all terms involving \(y\) to one side and all terms involving \(t\) to the other, we have set up the equation so that each side can be integrated independently. In our chemical reaction, the separation of variables led to the integral equation \(\int \frac{1}{y} dy = -0.6 \int dt\), which, upon integrating, gave us the natural logarithm of the substance's amount in relation to time. This method is especially useful when dealing with first-order reactions since it allows us to easily isolate the variable we are solving for (in this case, the quantity \(y\) of \(\delta\)-glucono lactone).

Initial Conditions

Initial conditions are values that define the state of a system at the starting point of analysis, which can be used to determine the constants in the general solution of a differential equation. Specifying initial conditions is necessary for obtaining a unique solution to a differential equation, as there are infinitely many functions that can satisfy the differential equation without them.

For the reaction in question, the initial condition is given as \(y(0) = 100\) grams, meaning that there were 100 grams of \(\delta\)-glucono lactone present at the start (\(t=0\)). Once we've separated the variables and integrated, we use this initial condition to find the value of the constant of integration \(C\). Doing so allows us to derive the specific equation \(y = 100e^{-0.6t}\) that accurately predicts the behavior of this chemical reaction. Without the initial condition, we could not have determined the constant \(C\), nor could we have made precise predictions about the substance's amount at different times.