Problem 23 - Chemistry: The Central Science

Question

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? Explain. (b) What are the reaction orders for A and B? What is the overall reaction order? (c) What are the units of the rate constant?

Step-by-Step Solution

Answer

(a) Doubling the concentration of A will not change the rate, as the rate is not dependent on the concentration of A. The rate constant will not change as it depends only on temperature. (b) The reaction order for A is 0, the reaction order for B is 2, and the overall reaction order is 2. (c) The units of the rate constant k are mol\(^{-1}\) L s\(^{-1}\).

Step 1: (a) How the rate changes when doubling the concentration of A

: The given rate law is: Rate \(= k[\mathrm{B}]^2\) In this expression, we can see that the rate is not dependent on the concentration of A. Therefore, when the concentration of A is doubled, the rate of the reaction will not change.

Step 2: (a) Will the rate constant change?

: The rate constant (k) is a proportionality constant that depends only on temperature, not on the concentration of reactants. Thus, when the concentration of A is doubled, the rate constant will not change.

Step 3: (b) Reaction orders for A and B

: For the reaction A + B → C, we have the rate law: Rate \(= k[\mathrm{B}]^2\) The reaction order with respect to A (m) and B (n) is given by the following equation: Rate \(= k[\mathrm{A}]^m[\mathrm{B}]^n\) Comparing the given rate law and the general rate law equation, we can see that: m = 0 (since the concentration of A is not present in the rate law) n = 2 (since the concentration of B is squared in the rate law) The reaction orders for A and B are 0 and 2, respectively.

Step 4: (b) Overall reaction order

: The overall reaction order is the sum of the individual reaction orders of A and B. So, in this case: Overall order = m + n = 0 + 2 = 2 The overall reaction order is 2.

Step 5: (c) Units of the rate constant

: Now, let's determine the units of the rate constant k. The rate of the reaction has units of concentration per time, typically represented as mol L\(^{-1}\) s\(^{-1}\). To obtain the units of k, we need to ensure that the dimensions on both sides of the rate law expression remain consistent. The rate law is: Rate \(= k[\mathrm{B}]^2\) Rearrange to isolate k: \(k = \dfrac{\text{Rate}}{[\mathrm{B}]^2}\) Now let's consider the units. The dimensions of the rate are mol L\(^{-1}\) s\(^{-1}\), and the dimensions of the concentration of B are mol L\(^{-1}\). Therefore, the units of k are: Units of k \(= \dfrac{\text{(mol L}^{-1}\text{ s}^{-1}\text{) }}{(\text{mol L}^{-1})^2}\) Units of k \(= \text{mol}^{-1} \text{L s}^{-1}\) The units of the rate constant k are mol\(^{-1}\) L s\(^{-1}\).

Key Concepts

Reaction Orders

Understanding reaction orders is key in determining how different reactants affect the speed of a chemical reaction. The reaction order signifies the power to which the concentration of a reactant is raised in the rate law equation.
In the exercise, we are given the rate law: \[ \text{Rate} = k[B]^2 \]This expression indicates that the reaction order with respect to reactant \( A \) is 0. This is because A does not appear in the rate law, meaning changes in its concentration do not affect the rate. For reactant \( B \), the reaction order is 2, as the concentration of \( B \) is squared in the rate law.

• Reaction order with respect to A: 0
• Reaction order with respect to B: 2

The overall reaction order is simply the sum of all the individual orders. In this example, the overall order is \( 0 + 2 = 2 \). Understanding individual and overall reaction orders helps in predicting how changes in concentrations will influence the rate of the reaction.

Rate Constant

The rate constant, denoted by \( k \), is a crucial component of the rate law. It acts as a proportionality factor that links the reaction rate to the concentrations of reactants following specific orders. Importantly, \( k \) is dependent solely on factors like temperature and the presence of a catalyst. It does not get influenced by the concentrations of reactants themselves.
In our example, even if the concentration of \( A \) is doubled, the rate constant \( k \) will remain unchanged. This highlights a vital characteristic of \( k \) - it is constant for a given reaction at a certain temperature. Understanding \( k \) assists in determining reaction rates under various conditions.

Units of Rate Constant

The units of the rate constant \( k \) give insight into the reaction order and the nature of the reaction process. It is determined by the dimensions required to balance the units of the rate expression.For our example, the rate has units of concentration per time, typically represented as mol L\(^{-1}\) s\(^{-1}\). As the concentration of \( B \) in the rate law is squared (\([B]^2\)), we find the units of \( k \) by rearranging:\[ k = \frac{\text{Rate}}{[\mathrm{B}]^2} \]Carrying out the dimensional analysis:

• The rate has units of mol L\(^{-1}\) s\(^{-1}\)
• The concentration of \( B \) has units of mol L\(^{-1}\)

Thus, the units of \( k \) are calculated as:\[ \text{Units of } k = \frac{\text{mol L}^{-1} \text{s}^{-1} }{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L s}^{-1} \]These units help confirm the reaction order and provide a consistency check for the derived rate law.