Calculate the value of Kp at 25 ∘C, where Kc = 40 A2 + B –> A2B
The Correct Answer and Explanation is:
To calculate the equilibrium constant ( K_p ) in terms of partial pressures from ( K_c ) (equilibrium constant in terms of concentration) for the given reaction:
[
\text{A}_2 + \text{B} \rightarrow \text{A}_2\text{B}
]
The relationship between ( K_p ) and ( K_c ) is given by the following equation:
[
K_p = K_c \left( RT \right)^{\Delta n}
]
Where:
- ( K_p ) is the equilibrium constant in terms of partial pressures.
- ( K_c ) is the equilibrium constant in terms of concentrations.
- ( R ) is the gas constant, which is 0.0821 L·atm/(mol·K).
- ( T ) is the temperature in Kelvin.
- ( \Delta n ) is the change in the number of moles of gas between products and reactants.
Step-by-Step Calculation:
1. Determine ( \Delta n ):
First, we need to find the change in the number of moles of gas during the reaction. From the given reaction:
[
\text{A}_2 (g) + \text{B} (g) \rightarrow \text{A}_2\text{B} (g)
]
- Moles of gas on the left (reactants): 1 mole of ( \text{A}_2 ) + 1 mole of ( \text{B} ) = 2 moles of gas.
- Moles of gas on the right (products): 1 mole of ( \text{A}_2\text{B} ) = 1 mole of gas.
Thus, ( \Delta n = 1 – 2 = -1 ).
2. Convert temperature to Kelvin:
Given that the temperature is 25°C, we convert it to Kelvin:
[
T = 25 + 273.15 = 298.15 \, \text{K}
]
3. Substitute values into the equation:
Now we can use the formula to calculate ( K_p ):
[
K_p = K_c \left( RT \right)^{\Delta n}
]
[
K_p = 40 \times (0.0821 \times 298.15)^{-1}
]
[
K_p = 40 \times (24.47)^{-1}
]
[
K_p = 40 \times 0.0409
]
[
K_p \approx 1.636
]
Final Answer:
The value of ( K_p ) at 25°C is approximately 1.636.
Explanation:
The calculation follows the formula that relates ( K_p ) and ( K_c ), which accounts for the change in the number of moles of gases and the temperature. The negative value of ( \Delta n ) indicates that there is a reduction in the number of moles of gas as the reaction progresses from reactants to products, affecting the relationship between ( K_p ) and ( K_c ). The decrease in moles of gas leads to a smaller value for ( K_p ) compared to ( K_c ).