Problem 91
Question
Under certain conditions, biphenyl, \(\mathrm{C}_{12} \mathrm{H}_{10},\) can be produced by the decomposition of cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12}\) \(2 \mathrm{C}_{6} \mathrm{H}_{12} \longrightarrow \mathrm{C}_{12} \mathrm{H}_{10}+7 \mathrm{H}_{2}\) This table represents part of the data obtained in the kinetics experiment. \begin{tabular}{lccc} \hline Time (s) & {\(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right](\mathrm{M})\)} & {\(\left[\mathrm{C}_{12} \mathrm{H}_{10}\right](\mathrm{M})\)} & {\(\left[\mathrm{H}_{2}\right](\mathrm{M})\)} \\ \hline 0.0 & 0.200 & 0.000 & 0.000 \\ 1.00 & 0.159 & 0.021 & \(-\) \\ 2.00 & 0.132 & \(-\) & \(-\) \\ 3.00 & \(-\) & 0.044 & \(-\) \\ \hline \end{tabular} (a) Fill in the missing concentrations. (b) Calculate the rate of reaction at \(1.5 \mathrm{~s}\).
Step-by-Step Solution
Verified Answer
(a) \([C_{12}H_{10}](2.00 \, s) = 0.034 \, M; [H_2](1.00 \, s) = 0.1435 \, M; [C_6H_{12}](3.00 \, s) = 0.112 \, M\). (b) Rate = 0.027 M/s.
1Step 1: Determine missing concentration of \([C_{12}H_{10}](2.00 \, s)\)
From the equation, each reaction of 2 moles of \(C_6H_{12}\) produces 1 mole of \(C_{12}H_{10}\). Initially, there was 0.200 M \(C_6H_{12}\), and now it is 0.132 M. Thus, 0.200 - 0.132 = 0.068 M \(C_6H_{12}\) has reacted. Therefore, at 2.00 seconds, \([C_{12}H_{10}] = \frac{0.068}{2} = 0.034 \, M\).
2Step 2: Determine missing concentration of \([H_2](1.00 \, s)\)
Using the stoichiometry, each reaction with 2 moles of \(C_6H_{12}\) forms 7 moles of \(H_2\). Thus, for the loss of 0.200 - 0.159 = 0.041 M \(C_6H_{12}\), the concentration of \(H_2\) is \(0.041 \times \frac{7}{2} = 0.1435 \, M\).
3Step 3: Determine missing concentration of \([C_6H_{12}](3.00 \, s)\)
At 3.00 seconds, \([C_{12}H_{10}] = 0.044 \, M\). Since every mole of \(C_{12}H_{10}\) formation involves 2 moles of \(C_6H_{12}\), the reacted \(C_6H_{12}\) is \(0.044 \times 2 = 0.088 \, M\). Thus, \([C_6H_{12}] = 0.200 - 0.088 = 0.112 \, M\).
4Step 4: Calculate the rate of reaction at \(1.5 \, s\)
To calculate the rate, use the change in concentration of \(C_6H_{12}\) over time. Between 1.00 s and 2.00 s, change in \([C_6H_{12}] = 0.159 - 0.132 = 0.027 \, M\). Time interval = 2.00 - 1.00 = 1.00 s. Therefore, rate = \(- \frac{0.027}{1.00} = -0.027 \, M/s\). The negative sign indicates a decrease in concentration; thus, the rate is 0.027 M/s.
Key Concepts
Reaction StoichiometryRate of ReactionConcentration Changes
Reaction Stoichiometry
Reaction stoichiometry is a fundamental concept in chemical kinetics. It deals with the quantitative relationships between reactants and products in a chemical reaction.
Stoichiometry allows us to determine how much of a reactant is needed to form a certain amount of product or vice-versa.
This concept is based on balanced chemical equations that show the proportions in which substances react.
For the decomposition of cyclohexane to form biphenyl, the chemical equation is:
\[2 \mathrm{C}_{6} \mathrm{H}_{12} \longrightarrow \mathrm{C}_{12} \mathrm{H}_{10}+7 \mathrm{H}_{2}\]
According to this equation, every 2 moles of \(\mathrm{C}_{6} \mathrm{H}_{12}\) decomposes to produce 1 mole of \(\mathrm{C}_{12} \mathrm{H}_{10}\) and 7 moles of \(\mathrm{H}_{2}\).
This proportionality is crucial to solving problems involving the calculation of unknown concentrations, as seen in the exercise, where the amounts of products like \(\mathrm{C}_{12} \mathrm{H}_{10}\) and \(\mathrm{H}_{2}\) were determined based on the reacted \(\mathrm{C}_{6} \mathrm{H}_{12}\).
The stoichiometric relationships guide these calculations by showing how changes in one substance affect the others.
Stoichiometry allows us to determine how much of a reactant is needed to form a certain amount of product or vice-versa.
This concept is based on balanced chemical equations that show the proportions in which substances react.
For the decomposition of cyclohexane to form biphenyl, the chemical equation is:
\[2 \mathrm{C}_{6} \mathrm{H}_{12} \longrightarrow \mathrm{C}_{12} \mathrm{H}_{10}+7 \mathrm{H}_{2}\]
According to this equation, every 2 moles of \(\mathrm{C}_{6} \mathrm{H}_{12}\) decomposes to produce 1 mole of \(\mathrm{C}_{12} \mathrm{H}_{10}\) and 7 moles of \(\mathrm{H}_{2}\).
This proportionality is crucial to solving problems involving the calculation of unknown concentrations, as seen in the exercise, where the amounts of products like \(\mathrm{C}_{12} \mathrm{H}_{10}\) and \(\mathrm{H}_{2}\) were determined based on the reacted \(\mathrm{C}_{6} \mathrm{H}_{12}\).
The stoichiometric relationships guide these calculations by showing how changes in one substance affect the others.
Rate of Reaction
The rate of reaction is another key concept in chemical kinetics. It refers to how quickly a reaction occurs.
This can be determined by measuring the change in concentration of a reactant or a product over a specified length of time.
The rate provides insight into the speed of the reaction and can be influenced by several factors like temperature, pressure, and concentration of reactants.
In the exercise, the rate of reaction for cyclohexane's decomposition was calculated over a time interval.
The change in concentration of \([\mathrm{C}_{6} \mathrm{H}_{12}]\) was measured between 1.00 s and 2.00 s, allowing us to find the rate of reaction at 1.5 s as \(0.027 \mathrm{M/s}\).
The rate is often given as a positive value, signifying the absolute change, while the negative sign indicates a decrease in concentration, typical for reactants.
Formally, it's represented as negative for reactants to emphasize the decrease, while the rate for products is often positive, emphasizing the increase.
This can be determined by measuring the change in concentration of a reactant or a product over a specified length of time.
The rate provides insight into the speed of the reaction and can be influenced by several factors like temperature, pressure, and concentration of reactants.
In the exercise, the rate of reaction for cyclohexane's decomposition was calculated over a time interval.
The change in concentration of \([\mathrm{C}_{6} \mathrm{H}_{12}]\) was measured between 1.00 s and 2.00 s, allowing us to find the rate of reaction at 1.5 s as \(0.027 \mathrm{M/s}\).
The rate is often given as a positive value, signifying the absolute change, while the negative sign indicates a decrease in concentration, typical for reactants.
Formally, it's represented as negative for reactants to emphasize the decrease, while the rate for products is often positive, emphasizing the increase.
Concentration Changes
Understanding concentration changes is essential in chemical kinetics as it directly relates to the progress of the reaction.
Concentration changes may be used to calculate reaction rates or to predict the remaining quantities of reactants and the amounts of products.
They are directly tied to reaction stoichiometry and specified using a balanced chemical equation.
In the provided exercise, the concentration changes were tracked over time to understand the decomposition of cyclohexane.
By measuring these changes at different intervals, like the reduction of \(\mathrm{C}_{6} \mathrm{H}_{12}\) or formation of \(\mathrm{C}_{12} \mathrm{H}_{10}\), one can accurately follow the reaction’s progress.
The calculations showed how the loss of \(\mathrm{C}_{6} \mathrm{H}_{12}\) aligns with the stoichiometric requirements to produce \(\mathrm{C}_{12} \mathrm{H}_{10}\) and \(\mathrm{H}_{2}\).
This illustrates how changes in concentration not only reflect the reaction's progress but also aid in confirming stoichiometric predictions.
Concentration changes may be used to calculate reaction rates or to predict the remaining quantities of reactants and the amounts of products.
They are directly tied to reaction stoichiometry and specified using a balanced chemical equation.
In the provided exercise, the concentration changes were tracked over time to understand the decomposition of cyclohexane.
By measuring these changes at different intervals, like the reduction of \(\mathrm{C}_{6} \mathrm{H}_{12}\) or formation of \(\mathrm{C}_{12} \mathrm{H}_{10}\), one can accurately follow the reaction’s progress.
The calculations showed how the loss of \(\mathrm{C}_{6} \mathrm{H}_{12}\) aligns with the stoichiometric requirements to produce \(\mathrm{C}_{12} \mathrm{H}_{10}\) and \(\mathrm{H}_{2}\).
This illustrates how changes in concentration not only reflect the reaction's progress but also aid in confirming stoichiometric predictions.
Other exercises in this chapter
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