Question

For the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g),\) the rate is \(0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s},\) when \([\mathrm{A}]_{0}=[\mathrm{B}]_{0}=1.0 \mathrm{~mol} / \mathrm{L}\). If the reaction is first order in \(\mathrm{B}\) and second order in \(\mathrm{A}\), what is the rate when \([\mathrm{A}]_{0}=\) \(2.0 \mathrm{~mol} / \mathrm{L}\) and \([\mathrm{B}]_{0}=3.0 \mathrm{~mol} / \mathrm{L} ?\)

Short answer

The rate is 2.4 mol/L·s.

Step-by-step solution

Write the rate law

For the given reaction, the rate law can be written based on the order of reaction with respect to reactants A and B. Since the reaction is first order in B and second order in A, the rate law is: \[ \text{Rate} = k[\text{A}]^2[\text{B}] \]

Identify known quantities

We are given the rate (0.20 mol/L·s) when \[ [\text{A}]_0 = 1.0 \text{ mol/L} \ [\text{B}]_0 = 1.0 \text{ mol/L} \]

Calculate the rate constant

Using the rate law and the given quantities, solve for the rate constant k: \[ 0.20 = k[1.0]^2[1.0] \ 0.20 = k \ k = 0.20 \text{ mol}^{-2}\text{L}^2\text{s}^{-1} \]

Substitute new concentrations into the rate law

Now substitute the new concentrations [\text{A}]_0 = 2.0 \text{ mol/L} and [\text{B}]_0 = 3.0 \text{ mol/L} into the rate law: \[ \text{Rate} = (0.20)[2.0]^2[3.0] \]

Calculate the new rate

Perform the calculations: \[ \text{Rate} = 0.20 \times 4 \times 3 = 2.4 \text{ mol/L·s} \]

Key concepts

Chemical Kinetics

Chemical kinetics is the branch of chemistry that deals with understanding the rates of chemical reactions. It focuses on how different conditions like concentration and temperature affect the speed at which a chemical reaction proceeds. This field is important because it helps us control reaction rates in various applications, from industrial processes to biological systems. By studying reaction rates, scientists can predict how long a reaction will take to complete, which can be vital in processes like drug manufacturing or food preservation.
In our given exercise, we are looking at a reaction involving gases A and B to form a product AB. Understanding the rate of this reaction helps us know how fast AB is being formed under different conditions.

Rate Law

The rate law is a mathematical equation that describes the relationship between the rate of a chemical reaction and the concentration of its reactants. It is usually written as: \[\text{Rate} = k[\text{A}]^m[\text{B}]^n\] where:

• \text{k is the rate constant
• \text{[A] and [B] are the concentrations of the reactants
• \text{m and n are the reaction orders with respect to A and B, respectively

In the given problem, the rate law is \[\text{Rate} = k[\text{A}]^2[\text{B}]\] since the reaction is second order in A (m=2) and first order in B (n=1). This means the rate of the reaction depends on the concentration of A squared and the concentration of B raised to the first power. Knowing the rate law allows us to predict how changes in concentrations of reactants affect the reaction rate. For instance, doubling the concentration of A will cause the rate to increase by a factor of four, while doubling the concentration of B will double the rate.

Reaction Order

Reaction order is an important concept in chemical kinetics that tells us how the rate of a reaction depends on the concentration of the reactants. It is determined experimentally and can be zero, first, second, or even fractional. In our exercise, the reaction is second order with respect to A and first order with respect to B. Here's what that means:

• Second order in A: Doubling the concentration of A (\text{[A]}) will lead to a fourfold increase in the rate because \(\text{Rate} \propto [\text{A}]^2\).
• First order in B: Doubling the concentration of B (\text{[B]}) will lead to a twofold increase in the rate because \(\text{Rate} \propto [\text{B}]\).

When the overall order is calculated, it is the sum of the exponents in the rate law. In this case, the overall reaction order is 3 (2 from A and 1 from B). Knowing the reaction order helps chemists understand the dynamics of the reaction and how to control it effectively.

Rate Constant

The rate constant (k) is a proportionality constant in the rate law that provides information about the speed of a reaction under specific conditions. It is determined experimentally and is unique for each reaction at a given temperature. The units of the rate constant vary depending on the overall order of the reaction. For our exercise, the rate constant is found using the initial conditions: \[\text{Rate} = k[\text{A}]^2[\text{B}]\]. Given \[\text{Rate} = 0.20 \text{ mol/L·s}\], \([\text{A}] = 1.0 \text{ mol/L}\), and \([\text{B}] = 1.0 \text{ mol/L}\), we can solve for k:

• 0.20 = k \times (1.0)^2 \times (1.0)
  0.20 = k
  Therefore, k = 0.20 \text{ mol}^{-2} \text{L}^2 \text{s}^{-1}

Knowing the rate constant, we can then predict the reaction rate for any given concentration of A and B by plugging the values into the rate law. In the final step, we applied k and the new concentrations to find: \[\text{Rate} = 0.20 \times (2.0)^2 \times (3.0) = 2.4 \text{ mol/L·s}\]. This shows how the rate constant plays a crucial role in determining how fast the reaction proceeds.