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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.

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 \( K_p = K_c (RT)^{\Delta n} \).

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.

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.