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.
See your post above.
To find the value of Kp for the given reaction, we need to relate Kp to Kc.
The relationship between Kp and Kc is given by the ideal gas law equation:
Kp = Kc * (RT)^(Δn)
Where:
Kp = equilibrium constant in terms of partial pressures
Kc = equilibrium constant in terms of molar concentrations
R = ideal gas constant (0.0821 L atm mol^(-1) K^(-1))
T = temperature in Kelvin
Δn = change in the number of moles of gas (products - reactants)
In the given reaction,
2NH3(g) --> N2(g) + 3H2(g)
The change in the number of moles of gas (Δn) is calculated as follows:
(1 + 3) - 2 = 2 moles of gas
We know that Kc = 4.0 x 10^(-2). The equilibrium constant, Kp, can be calculated using the equation mentioned above.
Substituting the values in the equation, we get:
Kp = (4.0 x 10^(-2)) * (0.0821 L atm mol^(-1) K^(-1))^(2)
Note that the temperature, T, is given as 1,000 K. Let's calculate the value of Kp.
Kp = (4.0 x 10^(-2)) * (0.0821 L atm mol^(-1) K^(-1))^(2)
Kp = 4.0 x 10^(-2) * (0.0821)^(2)
Kp = 4.0 x 10^(-2) * 0.00675961
Kp = 2.7038 x 10^(-4)
Therefore, the value of Kp for the given reaction is 2.7038 x 10^(-4).