Question

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2} .\) The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{s}^{-1} .\) Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from 1.00 \(\mathrm{L}\) of 0.600 \(\mathrm{MN}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of 20.0 \(\mathrm{h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

Short answer

The partial pressure of \(O_2\) produced from a \(1.00\,L\) solution of \(0.600\,M\, N_2O_5\) in chloroform at \(45^{\circ}C\) and decomposed over a period of \(20.0\,h\) in a \(10.0\,L\) container is given by the expression: \[P_{O_2} = \frac{([O_2] \times Volume)\times (0.0821\, \frac{L\cdot atm}{mol\cdot K} \times 318.15\,K)}{10.0\,L}\] where \([O_2] = \Delta[N_2O_5] \times \frac{1\,mol\,O_2}{2\,mol\,N_2O_5}\), and \(\Delta[N_2O_5] = [N_2O_5]_0 - \frac{[N_2O_5]_0}{e^{(1.0 \times 10^{-5}\mathrm{s}^{-1})(72000\,s)}}\).

Step-by-step solution

Write the balanced chemical equation for the decomposition of N₂O₅

The balanced chemical equation for the decomposition of N₂O₅ in chloroform is given by: \[2\, N_2O_5 \rightarrow 4\,NO_2 + O_2\] Dinitrogen pentoxide is decomposed to produce nitrogen dioxide and oxygen gas.

Use the integrated rate law for first-order reactions

For first-order reactions, the integrated rate law is given by: \[ln \frac{[N_2O_5]_0}{[N_2O_5]} = kt\] where: - \([N_2O_5]_0\) is the initial concentration of N₂O₅ - \([N_2O_5]\) is the concentration of N₂O₅ at time t - k is the rate constant, which is given as \(1.0 \times 10^{-5}\, s^{-1}\) - t is the reaction time in seconds First, we need to convert the reaction time from hours to seconds: \(20.0\,h \times \frac{3600\,s}{1\,h} = 72000\,s\) Now, we can rearrange the integrated rate law equation to solve for \([N_2O_5]\): \[[N_2O_5] = \frac{[N_2O_5]_0}{e^{kt}}\] Plug in the given values: \[[N_2O_5]= \frac{0.600\,M}{e^{(1.0 \times 10^{-5}\mathrm{s}^{-1})(72000\,s)}}\] Calculate the value of \([N_2O_5]\).

Calculate the change in concentration of N₂O₅

To find the change in concentration of N₂O₅, subtract the final concentration from the initial concentration: \(\Delta [N_2O_5] = [N_2O_5]_0 - [N_2O_5]\)

Calculate the concentration of O₂ produced

Use the stoichiometry from the balanced chemical equation to calculate the concentration of O₂ produced: \(\Delta[N_2O_5] \times \frac{1\,mol\,O_2}{2\,mol\,N_2O_5} = [O_2]\)

Calculate the moles of O₂ produced

Multiply the concentration of O₂ produced by the volume of the solution (in liters) to find the moles of O₂ produced: Moles of O₂ = \([O_2] \times Volume\)

Use the ideal gas law to find the partial pressure of O₂

The ideal gas law is given by: \(PV = nRT\) where: - P is the pressure (in atmospheres) - V is the volume (in liters), which is given as 10.0 L - n is the number of moles of gas - R is the ideal gas constant, which is \(0.0821\, \frac{L\cdot atm}{mol\cdot K}\) - T is the temperature in Kelvin First, convert the temperature from Celsius to Kelvin: \(T = 45°C + 273.15 = 318.15\,K\) Now, rearrange the ideal gas law to solve for the pressure (P): \(P = \frac{nRT}{V}\) Plug in the calculated moles of O₂ and given values: \[P_{O_2} = \frac{moles\,of\,O_2 \times 0.0821\, \frac{L\cdot atm}{mol\cdot K} \times 318.15\,K}{10.0\,L}\] Calculate the partial pressure of O₂ produced.

Key concepts

First-Order Reaction

First-order reactions are a key concept in chemical kinetics. These reactions have a rate that is directly proportional to the concentration of one reactant. For example, if the concentration of the reactant is doubled, the rate of the reaction is also doubled. This proportionality is expressed mathematically by the rate equation for first-order reactions:
\[ \text{Rate} = k[N_2O_5] \]
where \( k \) is the rate constant and \( [N_2O_5] \) is the concentration of the reactant, dinitrogen pentoxide. The characteristic feature of a first-order reaction is that its half-life—an important aspect in chemical kinetics, which is the time taken for the concentration of a reactant to fall to half its initial value—is constant and independent of its initial concentration.

Integrated Rate Law

Understanding the integrated rate law is key to solving problems related to reaction kinetics. For a first-order reaction, the integrated rate law relates the concentration of a reactant at any time to the initial concentration and the rate constant. It is expressed as:
\[ \text{ln} \frac{[N_2O_5]_0}{[N_2O_5]} = kt \]
This equation is derived from the differential rate law by integrating it over time. The natural logarithm (ln) of the ratio of initial concentration to the final concentration at time \( t \) is equal to the product of the rate constant and time. This integrated form is extremely useful when determining the amount of reactant remaining after a particular time, as it only requires the initial concentration, the rate constant and the elapsed time.

Partial Pressure of Gases

Partial pressure is a term used in chemistry to describe the pressure exerted by a single gas in a mixture of gases. It is proportional to the concentration of the gas in the mixture. In the case of a reaction occurring in a closed container, the partial pressure exerted by a particular gas produced in the reaction can be related back to the amount of the gas present. According to Dalton's law of partial pressures, the total pressure in a container is the sum of the partial pressures of individual gases. The partial pressure can be calculated using the ideal gas equation once the number of moles of the gas is known. This is crucial for calculating yields and understanding reaction dynamics in gaseous systems.

Ideal Gas Law

The ideal gas law is pivotal in linking together the physical properties of gases under ideal conditions. It is expressed as:
\[ PV = nRT \]
where \( P \) represents 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. The ideal gas law assumes that the particles of the gas have no volume and no intermolecular forces. It is used to predict the behavior of gases in reactions, like the decomposition of dinitrogen pentoxide, where it is essential for calculating the pressure exerted by a gas, which is proportional to the number of moles of the gas when the volume and temperature are constant. In kinetic exercises, this law allows for the derivation of the partial pressures witnessed as a consequence of chemical reactions.