Question

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

Short answer

Part (a): \ \( \ln \left( \frac{I}{I_0} \right) = -kx\)\ Part (b): \ \( \ln \left( \frac{V}{V_0} \right) = -kt\)

Step-by-step solution

- Represent Intensity as a Function

Let the intensity of light entering the solution be denoted as I_0 and the intensity of light leaving the solution after traveling a distance x be I. Assume that the fraction of light removed per unit length is a constant k.

- Express the Decrease in Intensity Differentially

Since the intensity of light decreases by a constant fraction per unit distance (k), the rate of change of the intensity I with respect to distance x is proportional to I. Mathematically, this can be represented as \ \(\frac{dI}{dx} = -kI\).

- Integrate the Differential Equation

To find the expression for I in terms of I_0, integrate \ \(\frac{dI}{I} = -k dx\). This yields \ \( \ln I = -kx + C \), where C is the integration constant.

- Determine the Integration Constant

When x=0, I = I_0 (the initial intensity entering the solution). Thus, \ \( \ln I_0 = C \). Substituting this back, we get \ \( \ln I = -kx + \ln I_0 \).

- Express in Required Form

Rewriting the equation from Step 4: \ \( \ln I - \ln I_0 = -kx \) \ \( \ln \left( \frac{I}{I_0} \right) = -kx \). This shows that the logarithm of the ratio of the intensity of light leaving to entering the solution is equal to the negative product of the fraction of light removed per unit length and the distance traveled in the solution.

- Represent Value as a Function

Let the initial value of savings be V_0 and the value remaining after time t be V. Assume that a constant fraction k of the value is lost per unit time due to inflation.

- Express the Decrease in Value Differentially

Since the value decreases by a constant fraction per unit time, the rate of change of the value V with respect to time t is proportional to V. Mathematically, this can be written as \ \(\frac{dV}{dt} = -kV\).

- Integrate the Differential Equation

To find the expression for V in terms of V_0, integrate \ \(\frac{dV}{V} = -k dt\). This yields \ \( \ln V = -kt + C \), where C is the integration constant.

- Determine the Integration Constant

When t=0, V = V_0 (the initial value of savings). Thus, \ \( \ln V_0 = C \). Substituting this back, we get \ \( \ln V = -kt + \ln V_0 \).

- Express in Required Form

Rewriting the equation from Step 4: \ \( \ln V - \ln V_0 = -kt \) \ \( \ln \left( \frac{V}{V_0} \right) = -kt \). This shows that the logarithm of the ratio of the remaining value to the initial value is equal to the negative product of the fraction lost per unit time and the savings time interval.

Key concepts

Rate Law

In chemistry and physics, a rate law expresses the relationship between the rate of a process and the concentration of the reactants. For a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant.

The general form for a first-order rate law is:
\[-\frac{d[A]}{dt} = k[A]\]

Here, \[k\] is the first-order rate constant, \[A\] is the concentration of the reactant, and \[-\frac{d[A]}{dt}\] represents the rate of decrease of the reactant concentration over time.

From this differential equation, you can see that as \[A\] decreases, the rate of reaction also decreases. This concept is useful in various fields, from chemical reactions to biological processes and even economics, where it describes phenomena like radioactive decay or inflation impact on savings.

Light Intensity

When light passes through a solution, it decreases in intensity. This decrease follows an exponential decay, which can be described by a first-order rate law.

According to Beer-Lambert Law, the decrease in intensity of light (I) as it travels through a solution is proportional to its initial intensity (\[I_0\]), the distance traveled (\[x\]), and a proportionality constant (\[k\]) related to the solution’s properties:
\[\ln \left(\frac{I}{I_0}\right) = -kx\]

Here, \[k\] represents the fraction of light absorbed per unit length. Thus, as light travels farther, its intensity decreases exponentially.

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They describe how a quantity changes over time or space.

For first-order rate laws, we often encounter differential equations such as:
\[\frac{dI}{dx} = -kI\]

and
\[\frac{dV}{dt} = -kV\]

These equations are solved by separation of variables and integration to find relationships between the variables. In each case, integrating results in a natural logarithm relationship, which can then be transformed into an exponential function.

Integral Calculus

Integral calculus involves calculating the integral of a function, which can represent areas under curves, accumulated quantities, and more. In first-order rate laws, integration helps us find the relationship between initial and final quantities.

When you integrate a first-order differential equation such as:
\[\int \frac{1}{I} \frac{dI}{dx} = -k \int dx\]

you find that:
\[\ln I = -kx + C\]

where \[C\] is a constant. By determining \[C\] using initial conditions (e.g., \[I_0\] at \[x=0\]), you can derive solutions that describe how the system evolves. This is central in applying first-order kinetics to real-world problems.