Question

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

Short answer

It takes approximately 2,403 seconds for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 g\). The volume of \(\mathrm{O}_{2}\) produced up to this point is approximately 0.384 L at 745 mmHg and \(45^{\circ}C\).

Step-by-step solution

Apply First-Order Reaction Formula

The integrated rate law for a first-order reaction is given by: \(\log[\mathrm{N}_{2} \mathrm{O}_{5}]_{t} = -kt + \log[\mathrm{N}_{2} \mathrm{O}_{5}]_{0}\). We need to calculate the time it takes for \(80.0 g\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to reduce to \(2.5 g\). The rate constant \(k\) is \(6.2 \times 10^{-4} s^{-1}\). Converting the masses to moles (molar mass of \(\mathrm{N}_{2} \mathrm{O}_{5} = 108 g/mol\)), we get \([\mathrm{N}_{2} \mathrm{O}_{5}]_{0} = 80.0/108\) mol and \([\mathrm{N}_{2} \mathrm{O}_{5}]_{t} = 2.5/108\) mol.

Solve for Time

Substitute the values into the integrated rate law and solve for \(t\). The obtained answer is the time it takes for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to reduce to 2.5 g.

Use the Ideal Gas Law

To find the volume of \(\mathrm{O}_{2}\) produced up to this point, use the ideal gas law: \(PV = nRT\). The pressure \(P\) can be converted to atm (745 mmHg = 0.98 atm). The temperature \(T\) should be in Kelvin, so convert \(45^{\circ}C\) to 318 K. The universal gas constant \(R\) is 0.0821 L·atm/K·mol. The number of moles \(n\) can be calculated as the difference between initial and final moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) divided by 2 (because the stoichiometry of the reaction shows that 1 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\) produces \(\frac{1}{2}\) mol \(\mathrm{O}_{2}\)).

Key concepts

Integrated Rate Law

The integrated rate law is a crucial tool when dealing with reactions that follow a first-order rate process. In a first-order reaction, the rate at which the concentration of a reactant decreases is directly proportional to its concentration at any given time. The mathematical expression of the integrated rate law for such reactions is: \ \( \log[\mathrm{N}_2 \mathrm{O}_5]_t = -kt + \log[\mathrm{N}_2 \mathrm{O}_5]_0 \).
Here, \([\mathrm{N}_2 \mathrm{O}_5]_t\) and \([\mathrm{N}_2 \mathrm{O}_5]_0\) are the concentrations at time \(t\) and the initial concentration, respectively. The rate constant \(k\) provides information about the speed of the reaction. For our exercise, this constant is \(6.2 \times 10^{-4} \mathrm{s}^{-1} \).
By inserting values derived from mass to moles conversion into the integrated rate law, one can solve for time \(t\). This calculation will help determine how long it takes for the concentration to decrease to the desired level. Employing the concept of integrated rate law is beneficial in real-world applications, such as predicting the shelf life of pharmaceuticals.

Ideal Gas Law

The ideal gas law is an equation of state that describes the behavior of ideal gases. It is expressed as \( PV = nRT \), where \(P\) denotes pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In the exercise, we want to determine the volume of \(\mathrm{O}_2\) gas produced. Using the ideal gas law requires careful unit conversion. Here, the pressure is converted from mmHg to atm (745 mmHg = 0.98 atm) and the temperature from Celsius to Kelvin (\(45^\circ C = 318\,K\)).
Knowing these conditions, you can insert them into the equation to solve for the gas's volume. The universal constant \(R\) is crucial in this calculation and is typically \(0.0821\,\text{L}\cdot\text{atm/K}\cdot\text{mol}\).

• Convert pressure to atm for calculations
• Ensure temperature is in Kelvin
• Plug all values into \( PV = nRT \) to find \(V\)

This law simplifies the analysis of gas production and is widely used across various scientific fields.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It follows the coefficients of a balanced chemical equation, offering insight into how reactants convert into products.
In this reaction, \(\mathrm{N}_2 \mathrm{O}_5 \) decomposes into \(\mathrm{N}_2 \mathrm{O}_4 \) and \(\frac{1}{2}\mathrm{O}_2 \). Stoichiometry indicates that each mole of \(\mathrm{N}_2 \mathrm{O}_5 \) produces half a mole of \(\mathrm{O}_2\). This is fundamental when calculating the amount of \(\mathrm{O}_2\) produced.
Knowing the moles of \(\mathrm{N}_2 \mathrm{O}_5\) initial and final allows you to determine the change in moles, which then aids in deducing the corresponding moles of produced \(\mathrm{O}_2\).
This entire stoichiometric analysis is essential for accurate determination of reaction outcomes and is critical for practical applications such as industrial synthesis and laboratory preparations.