# 4.4: Problems

- Page ID
- 106823

Problem \(\PageIndex{1}\)

In each case,

- Identify the dependent and independent variables.
- Determine if the differential equation is separable.
- Determine if the differential equation is linear.
- Find the general solution.
- Find the particular solution using the given initial condition.
- Verify that your solution satisfies the differential equation by substitution.

- \(\frac{dy}{dx}=\frac{y+2}{x-3}, y(0)=1\)
- \(x'=e^{x+t}, x'(0)=2\)
- \(\frac{dy}{dx}-\frac{3}{x}y=2x^4, y(1)=1\)
- \(\frac{df}{dt}=\frac{3t^2}{f}, f(2)=4\)
- \(h'(t)+2h(t)=4, h(0)=1\)

Problem \(\PageIndex{2}\)

Consider the reaction \(A \overset{k}{\rightarrow}B\). The rate of disappearance of A is proportional to the concentration of A, so:

\[-\frac{d[A]}{dt}=k[A] \nonumber\]

1) Obtain [A]\((t)\) and [B]\((t)\).

2) Using the definition of half-life \((t_{1/2})\), obtain an expression for \((t_{1/2})\) for this mechanism. Your result will be a function of \(k\).

3) Sketch \([A](t)\) and \([B](t)\) for the case \(k=0.1 s^{-1}\), \([B]_0=0\) and \([A]_0=10^{-3}M\). Remember that you are expected to do this without the help of a calculator.

Problem \(\PageIndex{3}\)

Consider the reaction \(2A \overset{k}{\rightarrow}B\). This mechanism is called a bi-molecular reaction, because the reaction involves the collision of two molecules of reactant. In this case, the rate of disappearance of A is proportional to the square of the concentration of A, so:

\[-\frac{1}{2}\frac{d[A]}{dt}=k[A]^2 \nonumber\]

Notice that the rate is proportional to the square of the concentration, so this is a second-order reaction.

Assume that the initial concentration of [A] is [A]\(_0\), and the initial concentration of [B] is zero.

- Obtain an expression for [A]\((t)\).
- Write down a mass balance (a relationship relating [A](t), [B](t), [A]\(_0\) and [B]\(_0\)) and obtain [B]\((t)\).
- Using the definition of half-life \((t_{1/2})\), obtain an expression for \((t_{1/2})\) for this mechanism. Your result will be a function of \(k\) and [A]\(_0\).

Problem \(\PageIndex{4}\)

Obtain \([A](t), [B](t),\) and\([C](t)\) for the following mechanism:

\[A \overset{k}{\rightarrow}B\overset{k}{\rightarrow}C \nonumber\]

Assume \([A](0)=[A]_0\), and \([B](0)=[C](0)=0\)

Note that this problem is identical to the one solved in Section 4.2 but with \(k_1=k_2\). Be sure you identify the step where the two problems become different.

Problem \(\PageIndex{5}\)

Consider the reaction

\[A \xrightleftharpoons[k_2]{k_1} B \nonumber\]

modeled mathematically by the following ODE

\[\frac{d[A]}{dt}=k_2[B]-k_1[A] \nonumber\]

The constants, \(k_1\) and \(k_2\) represent the kinetic constants in the forward and backward direction respectively, and [A] and [B] represent the molar concentration of A and B. Assume you start with initial concentrations \([A] (t = 0) = [A]_0\) and \([B](t = 0) = [B]_0\).

Mass conservation requires that \([A](t) + [B](t) = [A]_0 + [B]_0\)

- Obtain [A](t) and [B](t) in terms of \(k_1, k_2, [A]_0\) and \([B]_0\).
- Obtain expressions for the concentrations of A and B in equilibrium: \([A]_{eq} =[A](t \rightarrow \infty\)) and \([B]_{eq} =[B](t \rightarrow \infty\)).
- Prove that the equilibrium constant of the reaction, \(\frac{[B]_{eq}}{[A]_{eq}} = K_{eq}\), can be expressed as \[\frac{[B]_{eq}}{[A]_{eq}} = \frac{k_1}{k_2} \nonumber\]
- Assume that \(k_1=1 min^{-1}\), \(k_2=\frac{1}{2} min^{-1}\), \([A]_0=0\) and \([B]_0=0.1 M\), calculate \([A]_{eq}\) and \([B]_{eq}\), and sketch \([A](t)\) and \([B](t)\) to the best of your abilities.