Question

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, a \(A \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\) d. The half-life for a second-order reaction increases as the reaction proceeds.

Short answer

The true statement among the given options is statement d: The half-life for a second-order reaction increases as the reaction proceeds.

Step-by-step solution

Statement a: The half-life for a zero-order reaction increases as the reaction proceeds.

To determine the validity of this statement, we recall the expression for half-life for a zero-order reaction. This expression is \(t_{1/2} = \frac{[\mathrm{A}]_0}{2k}\), where \(t_{1/2}\) is the half-life, \(k\) is the rate constant, and \([\mathrm{A}]_0\) is the initial concentration. For a zero-order reaction, as the reaction progresses, the concentration of A decreases, but the half-life remains constant, since it depends only on the initial concentration and the rate constant. Therefore, statement a is false.

Statement b: A catalyst does not change the value of the rate constant.

A catalyst is a substance that increases the rate of a chemical reaction without undergoing any permanent change itself. It does this by providing an alternative reaction pathway with a lower activation energy. As a result, a catalyst increases the reaction rate by affecting the rate constant. So, statement b is false.

Statement c: The half-life for a reaction, a \(A \longrightarrow\) products, that is first-order in A increases with increasing \([\mathrm{A}]_{0}\).

For a first-order reaction, we have the half-life expression, \(t_{1/2} = \frac{0.693}{k}\), where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. As we can see, the half-life for a first-order reaction is not dependent on the initial concentration of A, thus it does not increase with increasing \([\mathrm{A}]_{0}\). Therefore, statement c is false.

Statement d: The half-life for a second-order reaction increases as the reaction proceeds.

For a second-order reaction, the half-life expression is \(t_{1/2} = \frac{1}{k[\mathrm{A}]_0}\), where \(t_{1/2}\) is the half-life, \(k\) is the rate constant, and \([\mathrm{A}]_0\) is the initial concentration. From this expression, we can see that the half-life depends inversely on the initial concentration. As the reaction progresses, the concentration of A decreases, so the half-life increases. Therefore, statement d is true. In conclusion, the true statement(s) among the given options is statement d: The half-life for a second-order reaction increases as the reaction proceeds.

Key concepts

Zero-Order Reaction

In a zero-order reaction, the rate of reaction is independent of the concentration of reactants. This means that the rate remains constant over time, no matter how much reactant is present. An everyday example could be taking a constant spoonful of ice cream until it runs out. The rate being constant implies that the graph of concentration versus time is a straight line with a negative slope. A key formula here is the half-life expression for zero-order reactions: \(t_{1/2} = \frac{[\text{A}]_0}{2k}\), where \(t_{1/2}\) is the half-life, \([\text{A}]_0\) is the initial concentration, and \(k\) is the rate constant.
The unique factor about zero-order reactions is that their half-life decreases as time goes on because it is directly proportional to the remaining concentration of the reactant. However, overall, the half-life depends on the initial concentration and not on how far the reaction has proceeded.

Catalyst Effect on Rate Constant

A catalyst is like a secret shortcut through a chemistry problem. It speeds up reactions by providing an alternate pathway with a lower activation energy. Think of it as having a fast pass at an amusement park ride. But unlike fast passes that might cost more, using a catalyst doesn't change the price or amount of reactants you need – it leaves them unchanged by the end.
However, don't be fooled into thinking that it doesn't affect the rate constant. It actually alters the rate constant \(k\), increasing it by offering this easier path for the reaction to proceed. This makes reactions faster in the presence of a catalyst but doesn't modify the overall thermodynamics or end result of the reaction.

First-Order Reaction

In first-order reactions, the rate of reaction is directly proportional to the concentration of one reactant. Imagine walking up stairs: each step is dependent on the last one taken. Such linear dependence makes first-order kinetics extremely straightforward.
The half-life for a first-order reaction is constant, described by the equation \(t_{1/2} = \frac{0.693}{k}\), where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. What stands out here is that the half-life remains unchanged regardless of how much or little of the substance you begin with. This concept is crucial for processes like radioactive decay and pharmacokinetics.

Second-Order Reaction

Second-order reactions depend on either the concentrations of two reactants or the square of the concentration of a single reactant. Picture two people shaking hands: if one person isn’t there, the handshaking stops. This interaction dependency means the rate changes significantly based on concentration changes.
The formula for half-life in second-order kinetics, \(t_{1/2} = \frac{1}{k[\text{A}]_0}\), shows a direct inverse relationship with initial concentration. As the reaction proceeds and the concentration drops, the half-life expands. This variable half-life makes the kinetics study more complex compared to zero or first-order reactions but provides richer information about molecular interactions.

Half-Life Calculation

The concept of half-life is integral to understanding reaction dynamics. It signifies the time required for half of the reactant to be consumed in a reaction. Imagine having a delicious pie and timing how long it takes until only half of it remains.
The beauty of half-life equations lies in how they change with the order of reactions. For zero-order, half-life decreases over time due to the constant consumption rate, making it dependent on initial concentration \([\text{A}]_0\). In contrast, first-order reactions boast a constant half-life, independent of how much you start with, thanks to their reliance on the rate constant \(k\). Finally, the second-order reaction's half-life lengthens as concentrations plummet, resembling a never-ending series of half-consumed pies. Each of these provides insight into the temporal progression of a reaction.