Problem 27 - Chemistry The Central Science In Si Units

Question

A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[A]^{3}\). (a) If [B] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

Step-by-Step Solution

Answer

(a) The rate stays the same, and the rate constant does not change. (b) Order of A: 3; B: 0; Overall: 3. (c) Units of k: \(\text{M}^{-2}\text{s}^{-1}\).

Step 1: Understanding the Rate Law

The given rate law for the reaction is \( \text{Rate} = k[A]^3 \). This indicates that the rate of reaction depends only on the concentration of \([A]\), raised to the third power.

Step 2: Analyzing the Effect of Doubling [B] on Rate

Since \([B]\) does not appear in the rate law, any change in its concentration, such as doubling \([B]\), will not affect the rate of the reaction. Thus, the rate remains unchanged. The rate constant, \(k\), also remains the same as it is a characteristic of the reaction under specific conditions of temperature and pressure.

Step 3: Determining Reaction Orders

The reaction order with respect to \(\mathrm{A}\) is 3, as its exponent in the rate law is 3. \(\mathrm{B}\) does not appear in the rate law, so its order is 0. Thus, the overall reaction order is the sum of individual orders: \(3+0=3\).

Step 4: Finding the Units of Rate Constant

The rate of reaction is expressed in units of concentration over time, typically \( \text{M/s} \). For the given rate law \( \text{Rate} = k[A]^3 \), rearranging gives \( k = \text{Rate}/[A]^3 \). Substituting the units: \( [k] = \text{M/s} / \text{M}^3 = \text{M}^{-2}\text{s}^{-1} \).

Key Concepts

Reaction Order

In chemical kinetics, the reaction order tells us how the concentration of reactants affects the rate of reaction. The given rate law \( \text{Rate} = k[A]^3 \) indicates the reaction order with respect to each reactant. Here, the reactant \([A]\) is raised to the third power, which means that the reaction is third order with respect to \(A\). This implies that the rate of the reaction is very sensitive to changes in the concentration of \(A\).Interestingly, the reactant \(B\) does not appear at all in the rate law equation, so it is considered to have a zero order. Changes in its concentration do not affect the reaction rate. Therefore, the overall reaction order is simply the sum of the orders concerning each reactant. In this case, \(3 + 0 = 3\).

• Third order with respect to \(A\)
• Zero order with respect to \(B\)
• Overall reaction order: 3

Rate Constant

The rate constant, \(k\), is a crucial part of the rate law equation \(\text{Rate} = k[A]^3 \). It provides key insights into the speed of a reaction. The rate constant is unique to each reaction and can differ under varying conditions of temperature and pressure. It captures how quickly a reaction can occur at a set of given conditions.It's important to note that the rate constant itself does not change when the concentrations of reactants, such as \(B\), change unless the reaction conditions change. So even if \([B]\) is doubled, the rate constant \(k\) remains constant.

• Characterizes the reaction speed
• Unique to each reaction
• Remains constant under set conditions

Units of Rate Constant

Determining the units of the rate constant, \(k\), is important as it reflects the reaction's order. For the rate law \(\text{Rate} = k[A]^3 \), understanding the units involves rearranging the equation to solve for \(k\): \[ k = \frac{\text{Rate}}{[A]^3} \]. The units of the rate are typically given as concentration per time, such as \( \text{M/s} \) (molarity per second). Since \([A]\) is raised to the third power, the units of \( [A]^3 \) are \( \text{M}^3 \). Thus, when calculating \([k]\), you have:\[ [k] = \frac{\text{M/s}}{\text{M}^3} = \text{M}^{-2}\text{s}^{-1} \]. These units indicate that as the reaction order increases, the units of \(k\) adapt to balance the rate equation, reaffirming the third-order nature of our reaction.

• Order affects the units of \(k\)
• Results from dividing by the product of concentrations
• For third order: \( \text{M}^{-2}\text{s}^{-1} \)