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Chapter 12: Problem 268

Set up the expression for the rate of a bimolecular reaction (a) when the two initial reactants are identical and (b) when two different reactants are used. For (a) find the concentration in terms of the initial concentration, the rate constant and time, assuming that the reactants combine in a one \(-\) to \(-\) one ratio. When in a reaction two molecules react to produce one or more resultants it is branded as a bimolecular or Second Order Reaction. State some examples of bimolecular reactions.

Short Answer

Expert verified
For a bimolecular reaction involving (a) identical reactants, the rate law is given as \(rate = k[A]^2\), and for (b) different reactants, the rate law is given as \(rate = k[A][B]\). In case (a), the concentration of A over time can be expressed as \([A] = \frac{[A]_0}{1+ k[A]_0 t}\). Some examples of bimolecular reactions include the hydrolysis of an ester, the reaction between nitric oxide and hydrogen, and the reaction between hydrogen and iodine.

Step by step solution

01

(a) Identical reactants

For a second-order reaction involving identical reactants, the general form of the rate law is: \(rate = k[A]^2\) where k is the rate constant, and [A] is the concentration of the reactant A.
02

(b) Different reactants

For a second-order reaction involving two different reactants, the general form of the rate law is: \(rate = k[A][B]\) where k is the rate constant, [A] and [B] are the concentrations of reactants A and B, respectively.
03

Concentration over time for identical reactants

To find the concentration [A] over time, we will integrate the rate expression for identical reactants: \(rate = -\frac{d[A]}{dt} = k[A]^2\) Separate the variables: \(\frac{d[A]}{[A]^2} = -k dt\) Now integrate both sides: \(\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k \int_{0}^{t} dt \) Where \([A]_0\) is the initial concentration of A and t is the time elapsed. After integration, we get: \(\frac{-1}{[A]}-\frac{-1}{[A]_0}=-kt\) Rearrange the equation to solve for [A]: \(\frac{1}{[A]}=\frac{1}{[A]_0} + kt\) \([A] = \frac{[A]_0}{1+ k[A]_0 t}\) This equation gives the concentration of A in terms of initial concentration, rate constant, and time elapsed for a bimolecular reaction with identical reactants.
04

Bimolecular reaction examples

Here are some examples of bimolecular reactions: 1. The hydrolysis of an ester: \(CH_3COOC_2H_5 + H_2O \rightarrow CH_3COOH + C_2H_5OH\) 2. The reaction between nitric oxide and hydrogen: \(2NO + 2H_2 \rightarrow N_2 + 2H_2O\) 3. The reaction between hydrogen and iodine: \(H_2 + I_2 \rightarrow 2HI\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law for Second-Order Reactions
Understanding the rate law for second-order reactions is pivotal when studying chemical kinetics, which pertains to the speed at which reactions occur. A bimolecular, or second-order, reaction means that the rate at which products form is proportional to the product of the concentrations of two reacting species, or the square of one species if it is a reaction involving identical reactants.

For case (a), involving identical reactants, the rate law is expressed as:
\[rate = k[A]^2\]
with 'k' being the reaction rate constant and '[A]' being the molar concentration of the reactant. This squared concentration suggests that the reaction鈥檚 rate will quadruple if the concentration of A doubles, highlighting the sensitivity of the reaction rate to concentration changes in second-order reactions.

For case (b), when we have two distinct reactants, the rate law takes a slightly different form:
\[rate = k[A][B]\]
Here, '[A]' and '[B]' are the concentrations of the two different reactants. Both these representations underpin the key characteristics of second-order kinetics: a direct and non-linear dependency on reactant concentrations.
Concentration Over Time
The concentration of reactants in a bimolecular reaction is not static but changes over time as the reactants interact to form products. Determining how the concentration varies gives insight into the reaction progress and helps in predicting the amount of reactants or products at any given moment.

For identical reactants in a second-order reaction, using calculus allows us to integrate the rate law to link concentration and time:
\[\frac{1}{[A]} = \frac{1}{[A]_0} + kt\]
This derived equation shows that the inverse of the concentration of reactant 'A' at any given time 't' is equal to the inverse of the initial concentration plus the product of the reaction rate constant 'k' and time 't'. As time progresses, the denominator of the resulting fraction increases, meaning that the reactant concentration [A] decreases over time. For students, visualizing this relationship as a graph can be incredibly helpful for understanding the decay of reactant concentration and the nature of second-order kinetics.
Reaction Rate Constant
The reaction rate constant 'k' is the proportionality factor in the rate law, and it provides crucial information about the reaction speed. For second-order reactions, 'k' has the units of \(\text{M}^{-1}\text{s}^{-1}\)

, indicating how the reaction rate will change per molar concentration of the reactant(s) per second. The rate constant is not affected by concentration but can vary with temperature and the presence of a catalyst.

In the equation for concentration over time derived from the rate of a second-order reaction with identical reactants:
\[ [A] = \frac{[A]_0}{1+ k[A]_0 t} \]
'k' plays a central role in determining how quickly the concentration of A decreases. A larger 'k' suggests a faster reaction, causing the concentration of A to diminish more swiftly over time. It is important for students to recognize that 'k' is an intrinsic property of the reaction at a given temperature, thus serving as a fundamental parameter to comprehend and predict the dynamics of the reaction under study.

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Most popular questions from this chapter

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