What is the Kp for the following reaction:
2NH3(g) --> N2(g) + 3H2(g)
This reaction is contained in 1.0 L container at 1,000 K and has reached equilibrium. (Kc = 4.0 x 10^-2). You have initially added 1,220 moles of NH3(g) in the flask.
Kp = Kc(RT)^delta n.
n = mols products - mols reactants = -2
To find the equilibrium constant Kp for the reaction, you need to understand the relationship between Kp and Kc.
Kp is the equilibrium constant expressed in terms of partial pressures, while Kc is the equilibrium constant expressed in terms of molar concentrations. The two constants are related by the equation:
Kp = Kc(RT)^(Δn)
Where:
- Kp is the equilibrium constant in terms of partial pressures
- Kc is the equilibrium constant in terms of molar concentrations
- R is the ideal gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
- T is the temperature in Kelvin
- Δn represents the change in the number of gaseous moles from reactants to products
In the given reaction, the change in moles (Δn) is:
Δn = (moles of products) - (moles of reactants)
Δn = (1 mole of N2 + 3 moles of H2) - (2 moles of NH3)
Δn = 1 + 3 - 2
Δn = 2
Since the given equilibrium constant, Kc, is 4.0 x 10^-2, and the value of Δn is 2, we can use the equation relating Kp and Kc to find the value of Kp.
Kp = Kc(RT)^(Δn)
Given:
- R = 0.0821 L·atm/(mol·K)
- T = 1,000 K
- Δn = 2
- Kc = 4.0 x 10^-2
Substituting these values into the equation:
Kp = (4.0 x 10^-2)(0.0821 L·atm/(mol·K))(1,000 K)^(2)
Now, calculate the value of Kp using the equation above.