Question

Chlorine reacts with chloroform according to the reaction given below: \(\mathrm{Cl}_{2}+\mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}\) When the initial concentration of \(\mathrm{Cl}_{2}\) is doubled the reaction rate increases by a factor of \(1.41\). What is the order of the reaction with respect to \(\mathrm{Cl}_{2} ?\) a. \(1 / 2\) \mathbf{b} . ~ \(-1 / 2\) c. \(-1\) d. 2

Short answer

The order of the reaction with respect to \(\mathrm{Cl}_{2}\) is \(1/2\).

Step-by-step solution

Understand the Rate Law Expression

In a chemical reaction, the rate law expresses the rate of reaction as a function of the concentration of the reactants, each raised to a power called the order of reaction. For the reaction \(\mathrm{Cl}_{2}+\mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}\), the general rate law can be written as: \[ \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \]where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of reaction with respect to \(\mathrm{Cl}_{2}\) and \(\mathrm{CHCl}_{3}\), respectively. We need to determine \(m\).

Analyze the Impact of Doubling Concentration

According to the problem, when the initial concentration of \(\mathrm{Cl}_{2}\) is doubled, the reaction rate increases by a factor of 1.41. Mathematically, this can be expressed as:\[ \frac{\text{New Rate}}{\text{Old Rate}} = \frac{k \cdot (2[\mathrm{Cl}_{2}])^m \cdot [\mathrm{CHCl}_{3}]^n}{k \cdot ([\mathrm{Cl}_{2}])^m \cdot [\mathrm{CHCl}_{3}]^n} = 1.41 \]This simplifies to:\[ 2^m = 1.41 \]

Solve the Equation for m

To find the value of \(m\), take the logarithm of both sides:\[ \log(2^m) = \log(1.41) \]By logarithmic property, this becomes:\[ m \cdot \log(2) = \log(1.41) \]Now, solve for \(m\):\[ m = \frac{\log(1.41)}{\log(2)} \approx \frac{0.1492}{0.3010} \approx 0.5 \]

Interpret the Result

The result \(m \approx 0.5\) indicates that the reaction is half-order with respect to \(\mathrm{Cl}_{2}\). From the given choices, option a (1/2) is correct, which matches our calculated value. This means that the rate of reaction changes by the square root of the ratio of change in concentration when \(\mathrm{Cl}_{2}\) is altered.

Key concepts

Reaction Order

Understanding reaction order is crucial in determining how changes in reactant concentrations affect the rate of a reaction. In the given chemical equation, \[ \mathrm{Cl}_{2} + \mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4} + \mathrm{HCl} \]we observe that the order of reaction with respect to chlorine (\( \mathrm{Cl}_{2} \)) is key to predicting how the reaction rate will respond to variations in its concentration.
In this case, the order of reaction with respect to \( \mathrm{Cl}_{2} \) is determined to be 0.5.
This tells us that the rate of reaction is proportional to the square root of the concentration of \( \mathrm{Cl}_{2} \), which is neither proportional to the concentration directly (first-order) nor constant (zero-order).
Understanding this helps in devising strategies to control reaction speeds in industrial and laboratory settings.

Rate Law

The rate law for a chemical reaction is a mathematical expression that relates the rate of reaction to the concentration of its reactants. For the reaction in question, the rate law can be given as:\[ \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \]Here, \( k \) is the rate constant, while \( m \) and \( n \) represent the reaction orders with respect to \( \mathrm{Cl}_{2} \) and \( \mathrm{CHCl}_{3} \) respectively. The coefficients \( m \) and \( n \) help us understand how sensitive the rate is to concentration changes of the reactants involved.

• When \( m \) is 0, the rate is independent of \( [\mathrm{Cl}_{2}] \).
• If \( m \) is 1, the rate changes directly with changes in \( [\mathrm{Cl}_{2}] \).
• In this problem, \( m \) was found to be 0.5, indicating a half-order with respect to \( \mathrm{Cl}_{2} \).

By exploring these coefficients, scientists can predict reaction behavior under varying conditions.

Chemical Kinetics

Chemical kinetics is the study of factors affecting the speed or rate of chemical reactions. It is key in understanding how different variables such as concentration, temperature, and presence of catalysts influence the rate at which a chemical reaction proceeds.
In this particular exercise, kinetics was used to identify how altering the concentration of \( \mathrm{Cl}_{2} \) impacts the reaction's reaction rate.
This involves analyzing how reactant particles collide and transform into products over time.

• Experimentally, doubling the concentration of \( \mathrm{Cl}_{2} \) increased the rate by a factor of 1.41.
• By establishing the relationship between reactant concentrations and reaction rates, we can design experiments and industrial processes with predictable outcomes.

Understanding chemical kinetics allows chemists to control reaction conditions, enhancing efficiency in chemical manufacturing and minimizing undesired side reactions.

Rate Constant

The rate constant, \( k \), is a crucial factor in the rate law equation. It indicates the inherent speed of a reaction when all reactants are at unit concentration.
It is a fixed value specific to a particular reaction at a given temperature.

• The rate constant does not change with concentration or time but can vary with temperature.
• In the rate law: \( \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \), \( k \) provides the proportionality needed to equate the rate to the concentration terms.
• A larger \( k \) means a faster reaction, while a smaller \( k \) suggests a slower process.

Understanding and calculating the rate constant helps in comparing different reactions and establishing which conditions optimize reaction speed in real-world applications.