Problem 96
Question
The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).
Step-by-Step Solution
Verified Answer
Additionally, how many moles of B are needed in a vessel with 1 mole of A to maintain the same rate?
Answer: The rate of the reaction with 2 moles of A and 4 moles of B is 16k. To maintain the same rate with 1 mole of A, you would need 16 moles of B.
1Step 1: (a) Writing the rate expression
The given reaction is second-order in A and first-order in B. Therefore, the rate expression can be written as:
$$\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]$$
2Step 2: (b) Finding the rate of the reaction in terms of k
In the given one-liter vessel, we have 2 moles of A (squares) and 4 moles of B (circles). Using the rate expression, we can find the rate in terms of k:
$$\text{rate} = k[\mathrm{A}]^2[\mathrm{B}] = k(2\,\text{mol})^2(4\,\text{mol}) = 16k$$
So, the rate of the reaction is \(16k\).
3Step 3: (c) Filling the one-liter vessel with an appropriate number of circles (B) for the same rate
Since we want to maintain the same rate as in part (b) and we have the same rate constant, k, we can set up an equation using the rate expression:
$$16k = k[\mathrm{A}]^2[\mathrm{B}]$$
In the new scenario, there are 1 mole of A (square) in the one-liter vessel. So, you need to find how many moles of B (circles) are required to keep the same rate:
$$16k = k(1\,\text{mol})^2[\mathrm{B}]$$
$$16k = k[\mathrm{B}]$$
Now, solve for the concentration of B:
$$[\mathrm{B}] = \frac{16k}{k}$$
$$[\mathrm{B}] = 16\,\text{mol}$$
Therefore, in order to maintain the same rate as in part (b), you need 16 circles (moles of B) in the one-liter vessel in the new scenario.
Key Concepts
Rate of ReactionRate ExpressionOrder of Reaction
Rate of Reaction
The rate of reaction refers to how quickly or slowly a chemical reaction occurs. Imagine it as the speed at which reactants are turned into products. This is crucial in various applications, from industrial manufacturing to biochemistry. The rate can be influenced by several factors, including concentration and temperature.
Understanding the reaction rate helps predict how long a reaction will take, which is important in scenarios like cooking or even in designing engines. For example, if you see a reaction equation, knowing the rate tells you how different conditions (like the amount of each chemical) might affect the time it takes to complete.
To measure the reaction rate, we often look at how the concentration of reactants or products changes over a specific time period. For more complex chemical reactions, especially those involving multiple steps, advanced calculations may be required, but the basic concept remains the same: it's all about speed.
Understanding the reaction rate helps predict how long a reaction will take, which is important in scenarios like cooking or even in designing engines. For example, if you see a reaction equation, knowing the rate tells you how different conditions (like the amount of each chemical) might affect the time it takes to complete.
To measure the reaction rate, we often look at how the concentration of reactants or products changes over a specific time period. For more complex chemical reactions, especially those involving multiple steps, advanced calculations may be required, but the basic concept remains the same: it's all about speed.
Rate Expression
The rate expression—also known as the rate law—is a mathematical equation that links the rate of a chemical reaction to the concentration of the reactants. These expressions are pivotal because they help us understand the dependency of reaction rates on reactant concentrations.
For the reaction between substances A and B, the rate expression might look like: \(\text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n\). The symbols \(m\) and \(n\) are the orders of reaction with respect to each reactant, while \(k\) is the rate constant, which is unique to every reaction.
This particular expression tells you how changing the concentration of the reactants will impact the speed of the reaction. In our exercise example, the reaction was second-order in \([\mathrm{A}]\) and first-order in \([\mathrm{B}]\). This means the concentration of \([\mathrm{A}]\) influences the rate more significantly than \([\mathrm{B}]\), making it a key factor in accelerating the reaction.
For the reaction between substances A and B, the rate expression might look like: \(\text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n\). The symbols \(m\) and \(n\) are the orders of reaction with respect to each reactant, while \(k\) is the rate constant, which is unique to every reaction.
This particular expression tells you how changing the concentration of the reactants will impact the speed of the reaction. In our exercise example, the reaction was second-order in \([\mathrm{A}]\) and first-order in \([\mathrm{B}]\). This means the concentration of \([\mathrm{A}]\) influences the rate more significantly than \([\mathrm{B}]\), making it a key factor in accelerating the reaction.
Order of Reaction
The order of reaction is a concept that helps describe the power to which the concentration of a reactant is raised in the rate expression. It's basically a way to understand how each reactant affects the reaction rate.
For instance, if we have a hypothetical reaction where substance A is involved and the reaction is first-order with respect to A, increasing its concentration doubles the reaction rate. In contrast, a second-order reaction with respect to A would quadruple the rate for the same increase in A's concentration.
Knowing the order of a reaction provides valuable insight into the reaction mechanism. It helps chemists figure out exactly how molecules are interacting during the reaction. In practice, the order of reaction is determined empirically, often through experiments where the reaction rate is measured under varying concentrations.
In our exercise, the reaction was second-order in A and first-order in B, illustrating different levels of involvement from each reactant in driving the reaction forward.
For instance, if we have a hypothetical reaction where substance A is involved and the reaction is first-order with respect to A, increasing its concentration doubles the reaction rate. In contrast, a second-order reaction with respect to A would quadruple the rate for the same increase in A's concentration.
Knowing the order of a reaction provides valuable insight into the reaction mechanism. It helps chemists figure out exactly how molecules are interacting during the reaction. In practice, the order of reaction is determined empirically, often through experiments where the reaction rate is measured under varying concentrations.
In our exercise, the reaction was second-order in A and first-order in B, illustrating different levels of involvement from each reactant in driving the reaction forward.
Other exercises in this chapter
Problem 87
Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta
View solution Problem 89
The greatest increase in the reaction rate for the reaction between \(\mathrm{A}\) and \(\mathrm{C}\), where rate \(=k[\mathrm{~A}]^{1 / 2}[\mathrm{C}]\), is ca
View solution Problem 97
For the reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}$$ the rate expression is rate \(=k[\mathrm{~A}][\mathrm{B}]\) (a) Given three test tubes, wi
View solution Problem 99
The gas-phase reaction between hydrogen and iodine $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ proceeds with a rate constant for
View solution