Question

For the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})\), the value of \(\mathrm{K}_{\mathrm{C}}\) at \(250^{\circ} \mathrm{C}\) is 26 . The valuc of \(\mathrm{K}_{\mathrm{p}}\) at this tempcrature will \(\mathrm{bc}\) (1) \(0.605\) (2) \(0.57\) (3) \(0.83\) (4) \(0.46\)

Short answer

The value of \(K_p\) is \(0.605\).

Step-by-step solution

- Understand the Relationship Between Kc and Kp

The relationship between the equilibrium constant in terms of concentration \(K_c\) and the equilibrium constant in terms of pressure \(K_p\) is given by the formula \[K_p = K_c (RT)^{\Delta n}\]. Here, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(abla n\) is the change in the number of moles of gas.

- Calculate the Change in Number of Moles (Δn)

For the given reaction \[\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons \text{PCl}_5(\text{g}) \], the change in number of moles of gases, \(\Delta n\), can be determined as follows: \[\Delta n = n_{\text{products}} - n_{\text{reactants}}\]. Here, we have 1 mole of product (PCl5) and 2 moles of reactants (PCl3 + Cl2), so \[\Delta n = 1 - 2 = -1.\]

- Convert Temperature to Kelvin

The provided temperature is \(250^{\circ} \text{C}\). Convert this to Kelvin by adding 273.15: \[T(K) = 250 + 273.15 = 523.15 \text{ K}\]

- Substitute Values into the Kp Formula

Now use the formula to calculate \(K_p\): \[K_p = K_c (RT)^{\Delta n}\]. Substitute the given values: \[K_c = 26, R = 0.0821 \text{ L·atm·K}^{-1} \text{·mol}^{-1}, T = 523.15 \text{ K}, \Delta n = -1\].

- Solve for Kp

Plug the values into the equation and solve: \[K_p = 26 \left(0.0821 \times 523.15\right)^{-1}\] \[K_p = 26 \left(42.95\right)^{-1}\] \[K_p = 26 \times \frac{1}{42.95}\] \[K_p \approx 0.605\]

Key concepts

Relationship between Kc and Kp

In chemical equilibrium, the equilibrium constant can be expressed in different ways. The equilibrium constant in terms of concentration is denoted as \( K_c \), while the equilibrium constant in terms of pressure is denoted as \( K_p \). The relationship between \( K_c \) and \( K_p \) is crucial for understanding how gases behave under different conditions.
The formula connecting them is: \[ K_p = K_c (RT)^{\Delta n} \].
Here, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) represents the change in the number of moles of gas. This equation ensures that we can convert between concentration-based and pressure-based equilibrium constants using temperature and the change in the number of moles.
This formula is essential for chemistry problems involving gas-phase reactions, as it allows us to compare and convert different constant values to better understand the system's behavior.

Change in Number of Moles (Δn)

The change in the number of moles, denoted as \( \Delta n \), is a key part of the relationship between \( K_c \) and \( K_p \). It is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in a balanced chemical equation.
For the reaction provided: \[ \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons \text{PCl}_5(\text{g}) \], we need to count the moles for both sides of the equation.

• Moles of reactants: 1 (PCl\textsubscript{3}) + 1 (Cl\textsubscript{2}) = 2 moles
• Moles of product: 1 mole (PCl\textsubscript{5})

Therefore, the change in the number of moles \( \Delta n \) is: \[ \Delta n = n_{\text{products}} - n_{\text{reactants}} \].
Substituting the values we get: \[ \Delta n = 1 - 2 = -1 \] This negative value indicates there are fewer moles of products than reactants, necessary information when applying the formula \( K_p = K_c (RT)^{\Delta n} \).

Temperature Conversion to Kelvin

In chemical equations involving gases, it is essential to use the temperature in Kelvin because the gas constant \( R \) is usually given in units that include Kelvin (K). To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
For the problem given, the temperature is \( 250^{\circ} \text{C} \). To convert this to Kelvin: \[ T(K) = 250 + 273.15 = 523.15 \text{ K} \] Using Kelvin ensures that temperature values are always positive, which is necessary for calculations involving gas laws and equilibrium constants.

Equilibrium Constant Calculation

Having understood all the necessary components, we can now calculate the equilibrium constant in terms of pressure \( K_p \).
Using the formula: \[ K_p = K_c (RT)^{\Delta n} \] and substituting the given values: \[ K_c = 26, R = 0.0821 \text{ L·atm·K}^{-1} \text{·mol}^{-1}, T = 523.15 \text{ K}, \Delta n = -1 \] we can proceed with the calculation.
First, calculate the expression within the bracket: \[ RT = 0.0821 \times 523.15 = 42.95 \] Then, acknowledging that \( \Delta n = -1 \), we take the inverse of 42.95:
\( (RT)^{-1} = (42.95)^{-1} = \frac{1}{42.95}\approx0.0233 \) Finally, substitute this value back into the \( K_p \) equation:
\[ K_p = 26 \times 0.0233 \approx 0.605 \] Therefore, the value of \( K_p \) at 250°C is approximately 0.605.