Problem 82
Question
Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]
Step-by-Step Solution
Verified Answer
After following the steps, the value of \(K_P\) calculated should be the final answer.
1Step 1: Calculate Initial Pressure
The first step is to calculate the initial pressure of the iodine gas before dissociation. This can be done using the ideal gas law \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. The number of moles n can be calculated by dividing the mass of iodine by its molar mass. The initial pressure can then be calculated by rearranging the ideal gas law to solve for P: \[ P = \frac{nRT}{V} \]Plug in the values \(n = \frac{1~g}{253.8~g/mol}\), \(R = 0.0821~L.atm/mol.K\), \(T = 1200 + 273.15 = 1473.15~K\), and \(V = 500~mL = 0.5~L\).
2Step 2: Compute Degree of Dissociation
Next, calculate the degree of dissociation \(\alpha\) using the hint provided in the exercise. This can be achieved by calculating the ratio of observed pressure over calculated pressure.\[\alpha = \frac{P_{observed}}{P_{initial}}\]Substitute the given observed pressure \(1.51~atm\) and previously calculated initial pressure from step 1.
3Step 3: Use Degree of Dissociation to Find Partial Pressures
Now calculate the partial pressures at equilibrium for \(\mathrm{I}_2\) and \(\mathrm{I}\). This can be done by using the equation \(P_{total} = P_{\mathrm{I}_2} + 2P_{\mathrm{I}}\). The pressure of \(I_2\) is \((1-\alpha) \times P_{initial}\) and the pressure of \(\mathrm{I}\) is \(2 \times \alpha \times P_{initial}\). Substitute the values of \(\alpha\) and \(P_{initial}\) obtained from previous steps.
4Step 4: Calculate \(K_P\)
Finally, \(K_P\) can be calculated using the expression:\[ K_P = \frac{P_{I}^2}{P_{I_2}} \]Substitute the values of \(P_I\) and \(P_{I_2}\) obtained from step 3 to find the answer.
Key Concepts
DissociationIdeal Gas LawPartial PressureDegree of Dissociation
Dissociation
In chemistry, dissociation refers to the breaking down of a compound into its simpler molecules, atoms, or ions. This process occurs when a molecule, such as iodine gas represented by \( ext{I}_2\), splits into smaller particles—in this case, individual iodine atoms \( ext{I}\).
This phenomenon is essential in understanding reactions where initial compounds break down to form new substances. Dissociation can be a reversible process, which means the separated molecules (or atoms) can recombine to form the original compound under certain conditions.
In the context of our exercise, iodine dissociates when heated, and this process can be influenced by factors such as temperature and pressure. At a specific temperature, an equilibrium is established between the dissociated and undissociated iodine molecules.
This phenomenon is essential in understanding reactions where initial compounds break down to form new substances. Dissociation can be a reversible process, which means the separated molecules (or atoms) can recombine to form the original compound under certain conditions.
In the context of our exercise, iodine dissociates when heated, and this process can be influenced by factors such as temperature and pressure. At a specific temperature, an equilibrium is established between the dissociated and undissociated iodine molecules.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that models the behavior of gases. It is represented as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
This law allows us to calculate any one of these variables if the other three are known, assuming the gas behaves ideally, which means no intermolecular forces and negligible volume of gas molecules.
In the original exercise, the ideal gas law is used to find the initial pressure of iodine gas before it begins to dissociate. This is crucial for further calculations of equilibrium constants, as it sets the baseline pressure against which the effects of dissociation are measured.
This law allows us to calculate any one of these variables if the other three are known, assuming the gas behaves ideally, which means no intermolecular forces and negligible volume of gas molecules.
In the original exercise, the ideal gas law is used to find the initial pressure of iodine gas before it begins to dissociate. This is crucial for further calculations of equilibrium constants, as it sets the baseline pressure against which the effects of dissociation are measured.
Partial Pressure
Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It is an important concept when dealing with reactions involving gases. In a closed system, the total pressure is the sum of the partial pressures of all gases present.
For the iodine dissociation exercise, the final equilibrium state includes both iodine molecules \( ext{I}_2\) and atoms \( ext{I}\). Therefore, the total pressure at equilibrium is the sum of the partial pressures of undissociated iodine molecules and the iodine atoms formed by dissociation.
Understanding partial pressures allows us to calculate the equilibrium constant \( K_P \) for the distribution of iodine atoms and molecules, which relies on knowing these pressures at equilibrium.
For the iodine dissociation exercise, the final equilibrium state includes both iodine molecules \( ext{I}_2\) and atoms \( ext{I}\). Therefore, the total pressure at equilibrium is the sum of the partial pressures of undissociated iodine molecules and the iodine atoms formed by dissociation.
Understanding partial pressures allows us to calculate the equilibrium constant \( K_P \) for the distribution of iodine atoms and molecules, which relies on knowing these pressures at equilibrium.
Degree of Dissociation
The degree of dissociation, represented by \( \alpha \), indicates the fraction of the original compound that has dissociated into its components. It is a measure of the extent of dissociation and varies with different conditions such as pressure and temperature.
In calculations, \( \alpha \) is found by comparing the observed pressure with the calculated initial pressure assuming no dissociation. It is defined by \( \alpha = \frac{P_{observed}}{P_{initial}} \), where \( P_{observed} \) is the actual measured pressure at equilibrium, and \( P_{initial} \) is what the pressure would be without dissociation.
The degree of dissociation helps in determining the equilibrium concentrations or pressures of the different species in a mixture, enabling the calculation of crucial quantities like the equilibrium constant \( K_P \).
In calculations, \( \alpha \) is found by comparing the observed pressure with the calculated initial pressure assuming no dissociation. It is defined by \( \alpha = \frac{P_{observed}}{P_{initial}} \), where \( P_{observed} \) is the actual measured pressure at equilibrium, and \( P_{initial} \) is what the pressure would be without dissociation.
The degree of dissociation helps in determining the equilibrium concentrations or pressures of the different species in a mixture, enabling the calculation of crucial quantities like the equilibrium constant \( K_P \).
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