The equilibrium constant KP for the following reaction is 4 atm^2 at 300 K. AB(s) ---> A(g) + B(g)
What is the equilibrium pressure of A(g) and B(g) above AB(s) at 300 K? What is KC at 300K?
Kp = pA*pB = 4.
I would let pA = p which means pB = p and solve for p.
Kp = Kc(RT)delta n
Could you please explain what you wrote above? Like can you replace all the constants with the numbers given so I get what you are saying...Thanks! Also how will I find the equilibrium pressure and Kc?
And how do I solve for p? I'm confused :(
Kp = 4 = partial pressure of A x partial pressure of B
4=p*p
4 = p^2
p=2 atm
partial pressure A = 2 atm at eqilibrium.
partial pressure of B = 2 atm at equilibrium.
Kp = Kc(RT)delta n
You know Kp and you know R and T. You can calculate delta n (it's mols reactants - moles products = 2-0=2)
Note that Kp = pA*pB and does NOT (repeat NOT) include AB since it is a solid.
So is this right:
4=Kc*((0.08206)(300K)^2)= .006600
Also what is the answer to this part of the question:
What is the equilibrium pressure of A(g) and B(g) above AB(s) at 300 K?
Kc is right but you worked it wrong.
It's 4 = Kc*(0.08206*300)^2
What do you not understand about partial pressure? Do you know what that means? I have posted the equilibrium partial pressures for both A and B in all of my responses but you continue to ask what is the answer.
Ohh I'm sorry I didn't see that! Sorry for the mishap :( I really appreciate your help!
You're welcome.
Hey, I just wanted to help and point out that delta n = moles of gaseous products - moles of gaseous reactants according to my chem lecture.
To determine the equilibrium pressure of A(g) and B(g) above AB(s) at 300 K, we need to consider the equilibrium constant KP and the stoichiometry of the reaction. The equilibrium constant KP is defined as the ratio of the partial pressures of the products (A(g) and B(g)) to the partial pressure of the reactant (AB(s)), each raised to the power of its stoichiometric coefficient.
The balanced equation for the reaction is:
AB(s) -> A(g) + B(g)
From the equation, we can see that the stoichiometric coefficient of AB(s) is 1, whereas for A(g) and B(g) it is also 1. This means that the equilibrium constant KP is equal to:
KP = (P_A * P_B) / P_AB^2
where P_A is the partial pressure of A(g), P_B is the partial pressure of B(g), and P_AB is the partial pressure of AB(s).
Given that KP = 4 atm^2, we can now solve for the equilibrium pressure of A(g) and B(g) in terms of P_AB.
KP = (P_A * P_B) / P_AB^2
Rearranging the equation, we get:
4 = (P_A * P_B) / P_AB^2
Cross multiplying, we have:
4 * P_AB^2 = P_A * P_B
Since we don't have any further information about the pressures of A(g), B(g), or AB(s), we cannot determine the specific values for P_A, P_B, or P_AB. However, we know that the two partial pressures must be related in such a way that their product is equal to 4 times the square of P_AB.
Regarding the equilibrium constant KC at 300 K, we can convert KP to KC using the equation:
KP = KC * (RT)^Δn
where R is the ideal gas constant (0.0821 atm/mol/K), T is the temperature in Kelvin (300 K), and Δn represents the change in the number of moles of gas between the products and the reactant.
Since the reaction involves a solid (AB(s)) and gases (A(g) and B(g)), Δn = (1 + 1) - 1 = 1.
Substituting the values, we have:
4 = KC * (0.0821 * 300)^1
Simplifying, we get:
4 = KC * 24.63
Dividing both sides by 24.63, we find:
KC = 4 / 24.63
Calculating this value gives us the equilibrium constant KC at 300 K, which is approximately 0.162 atm.