Question

The first-order rate constant for the decomposition of dimethyl ether: $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ is \(3.2 \times 10^{-4} \mathrm{~s}^{-1}\) at \(450^{\circ} \mathrm{C}\). The reaction is carried out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is \(0.350 \mathrm{~atm} .\) What is the pressure of the system after 8.0 min? Assume ideal behavior.

Short answer

The pressure after 8 minutes is 0.450 atm.

Step-by-step solution

Understand the Problem

We are given the decomposition reaction of dimethyl ether with a rate constant of \( k = 3.2 \times 10^{-4} \ \mathrm{s}^{-1} \) at \( 450^{\circ} \mathrm{C} \). The initial pressure is \( 0.350 \ \mathrm{atm} \), and we need to find the pressure after 8 minutes, assuming ideal gas behavior.

Convert Time Units

Convert the time from minutes to seconds because the rate constant is given in seconds. \[ 8.0 \ \text{min} = 480 \ \text{s} \]

Use First-Order Kinetics Equation

For a first-order reaction, the integrated rate law is: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] where \([A]_0\) is the initial concentration (pressure) and \([A]\) is the concentration at time \( t \).

Calculate Remaining Dimethyl Ether Pressure

Substitute the known values into the first-order equation to find the remaining concentration (pressure) of dimethyl ether:\[ \ln \left( \frac{0.350 \ \text{atm}}{P_{\text{CH}_3\text{OCH}_3}} \right) = (3.2 \times 10^{-4} \ \text{s}^{-1})(480 \ \text{s}) \]Calculate the term on the right-hand side: \[ \ln \left( \frac{0.350}{P_{\text{CH}_3\text{OCH}_3}} \right) = 0.1536 \]We solve for \(P_{\text{CH}_3\text{OCH}_3}\): \[ \frac{0.350}{P_{\text{CH}_3\text{OCH}_3}} = e^{0.1536} \approx 1.166 \]\[ P_{\text{CH}_3\text{OCH}_3} \approx \frac{0.350}{1.166} = 0.300 \ \text{atm} \]

Calculate Total Pressure After Reaction

Since 1 mole of dimethyl ether produces 3 moles of gas, use stoichiometry to find the total pressure. Initially, dimethyl ether contributes \( 0.350 \ \text{atm} \) (all gas present). The change in pressure due to decomposition into 3 times the amount of gas will be:\[ \Delta P = 0.350 - 0.300 = 0.050 \ \text{atm} \text{ (as decomposed)} \]The total pressure will be: \[ P_{\text{total}} = P_{\text{CH}_3\text{OCH}_3} + 3 \times \Delta P \]\[ P_{\text{total}} = 0.300 + 3 \times 0.050 = 0.450 \ \text{atm} \]

Ideal Gas Behavior Confirmation

Since we assumed ideal gas behavior, we use the ideal gas law concept of partial pressures: The total pressure is a sum of the pressures of each gas involved, which we summed considering the stoichiometric ratios of formation, and got \(0.450\ \mathrm{atm}\).

Key concepts

Rate Constant

The rate constant in a chemical reaction is a crucial parameter in defining the speed at which a reaction occurs.
For first-order reactions, like the decomposition of dimethyl ether here, the rate constant (\( k \)) determines how quickly the reactant, dimethyl ether in this case, breaks down into its products.
This constant can vary with temperature and is given at a specific temperature in the problem statement: \( 3.2 \times 10^{-4} \ \text{s}^{-1} \) at \( 450^{\circ} \text{C} \).
Understanding how the rate constant influences reaction progress helps us predict the extent of reaction at any given time.

• First-order reaction means that the rate of reaction is directly proportional to the concentration of one reactant.
• Large values of the rate constant signify a faster reaction while smaller values indicate a slower reaction.
• Remember, this constant is specific to a given temperature and reaction setup.

Integrated Rate Law

The integrated rate law allows us to calculate the concentration of a reactant at any time after the start of the reaction.
For first-order reactions, the integrated rate law equation is given by:\[\ln \left( \frac{[A]_0}{[A]} \right) = kt\]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant.
Using this equation, it becomes possible to determine how much of the original substance remains at a specific time.

• To use the integrated rate law, you must first convert any time units to match those of the rate constant, such as converting minutes to seconds.
• Substitute known values to solve for the unknown concentration following the time of interest.
• The exponential nature of the law helps us understand the reaction's progress over time.

Stoichiometry

Stoichiometry involves using the relationships or ratios of the quantities of reactants and products involved in a chemical reaction.
In the given reaction, dimethyl ether decomposes to form methane, hydrogen, and carbon monoxide in a 1:3 mole ratio.
This implies that for every mole of dimethyl ether decomposed, three moles of gaseous products are formed.Following the equation:\[\left(\text{CH}_{3}\right)_{2} \text{O}(g) \longrightarrow \text{CH}_{4}(g) + \text{H}_{2}(g) + \text{CO}(g)\]we see that decomposing dimethyl ether increases total pressure as three times more moles are produced.

• The stoichiometry gives a clear ratio: 1 mole of ether produces 3 moles of products.
• By examining pressure changes, we can calculate how much pressure is due to product formation.
• This scaled formation affects the system's pressure in measurable ways.

Ideal Gas Law

The ideal gas law is an equation of state for a hypothetical ideal gas and is an excellent approximation for many gases under a range of conditions.
It is given in the form:\[PV = nRT\]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
For constant-volume reactions, like in this exercise, changes in pressure directly relate to changes in the number of moles of gas.
Since we assume ideal behavior, the total pressure considers partial pressures of all gases: each gas contributes independently to the total pressure according to its mole counts.

• Ideal gas law assumptions are often valid when temperature is not too low, and pressure is not too high.
• Total gas pressure in this reaction sums contributions from all products and reactants.
• Behavioral assumptions simplify calculations by treating gases as if they have no interactions.