Consider the following reaction:
Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm, and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions
Kp=1.6
Kp=1.0*10^-4
Kp=1.2*10^5
I need to get this question immediately!! I've tried to figure this out for hours now. I've tried multiple ICE tables. Once I know how to do the first one I'm sure I can get the rest.
What's the reaction equation?
Oops! I thought that was there. It is A(g)<-> 2B(g)
...........A ==> 2B
initial...0.....1.0
change....+p....-p
equil.....p.....1.0-p
Check that.
so this set up would be 1.6= (1-p)^2/p
which then turns to p^2+1.6p-1 then solve the quadratic equation?
To solve this problem, we can utilize the concept of equilibrium expressions and the relationship between the concentrations (or partial pressures) of reactants and products.
The equilibrium expression for the given reaction is defined as follows:
Kp = (PA^x * PB^y) / (PC^z)
where PA, PB, and PC represent the partial pressures of A, B, and C, respectively, and x, y, and z are the stoichiometric coefficients of the balanced equation.
In this case, the balanced equation is not provided, so let us assume it to be:
A + B ⇌ C
Now, let's solve the problem step by step for each value of Kp:
1. For Kp = 1.6:
Given the initial partial pressure of B is 1.0 atm and A is 0.0 atm.
Let's represent the equilibrium partial pressures as follows:
PA = x
PB = 1.0
PC = y
Using the equilibrium expression, we have:
1.6 = (x * 1.0) / y
Since the initial partial pressure of A is 0.0 atm, the equilibrium partial pressure of A will also be 0.0 atm. Therefore, x = 0.
Substituting the values, we have:
1.6 = (0 * 1.0) / y
1.6 = 0 / y
Since the numerator is 0, the value of y can be any non-zero number. However, to simplify the problem, we can assume y = 1.
Thus, the equilibrium partial pressures of A and B are PA = 0.0 atm and PB = 1.0 atm.
2. For Kp = 1.0 * 10^-4:
Following the same steps as above, we have:
(1.0 * 10^-4) = (x * 1.0) / y
Assuming x = 0 and y = 1, we get:
(1.0 * 10^-4) = (0 * 1.0) / 1
(1.0 * 10^-4) = 0
Since the left side of the equation is nonzero, there is no value of PA that can satisfy this equilibrium condition. So, the equilibrium partial pressure of A will be 0.0 atm, and the equilibrium partial pressure of B will be 1.0 atm.
3. For Kp = 1.2 * 10^5:
Following the same steps, we have:
(1.2 * 10^5) = (x * 1.0) / y
Assuming x = 0 and y = 1, we get:
(1.2 * 10^5) = (0 * 1.0) / 1
(1.2 * 10^5) = 0
Similarly to the previous case, there is no value of PA that can satisfy this equilibrium condition. So, the equilibrium partial pressure of A will be 0.0 atm, and the equilibrium partial pressure of B will be 1.0 atm.
In summary, the equilibrium partial pressures of A and B for each value of Kp are:
1. For Kp = 1.6: PA = 0.0 atm, PB = 1.0 atm
2. For Kp = 1.0 * 10^-4: PA = 0.0 atm, PB = 1.0 atm
3. For Kp = 1.2 * 10^5: PA = 0.0 atm, PB = 1.0 atm
Please note that this explanation assumes the given reaction stoichiometry, and the values of x, y, and z were assumed as 0 and 1 based on the given initial partial pressures. The actual values may vary based on the actual balanced equation and stoichiometric coefficients.