Some SO3(g), intially at 1.000 atm, is placed in an empty reaction vessel. At equilibrium, the total pressure is 1.017 atm. Assuming the partial pressures are in atmospheres, what is Kp for the reaction below?
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
............2 SO3(g) ⇌ 2 SO2(g) + O2(g)
I...............1.................0.................0
C............-2x................2x...............x
E.............1-2x..............2x...............x
Ptotal = 1.017 = 1-2x+2x+x = 1+x so x = 0.017 atm.
Now under the E line just fill in as follows:
E.............1-2x..............2x...............x
.............1-2*0.017........2*0.017........0.017
Then plug those pressures into Kp expression and solve for Kp.
Post your work if you get stuck.
BTW, the last time I talked with Kevin s/he promised the same screen name would be used. It helps us help you if you stuck to one screen name..
Well, it seems like SO3 had a little bit of stage fright when it entered the reaction vessel, because its pressure slightly increased from 1.000 atm to 1.017 atm. Must have been nervous!
Now, let's find out what our good friend Kp is for this reaction. Remember, Kp measures the ratio of the partial pressure of the products to the partial pressure of the reactants. In this case, we have:
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
Since we know the total pressure at equilibrium is 1.017 atm, we can subtract the partial pressure of SO3 from that to find the sum of the partial pressures of SO2 and O2. So:
1.017 atm - x = 2x
Here, x represents the partial pressure of SO3 (in atm) at equilibrium. So, solving for x, we have:
1.017 atm = 3x
x = 1.017 atm / 3
x ≈ 0.339 atm
Now, knowing the partial pressure of SO3 at equilibrium, we can calculate Kp:
Kp = (P(SO2))^2 * P(O2) / (P(SO3))^2
Kp = (2 * x)^2 * x / (x)^2
Kp = (2 * 0.339 atm)^2 * 0.339 atm / (0.339 atm)^2
Kp ≈ 0.231
So, Kp ≈ 0.231. Just remember, it's normal for reactions to have a little bit of pressure to perform!
To find the value of Kp for the reaction, we can use the equilibrium partial pressures of the gases.
Let's assume that the equilibrium partial pressures for SO3, SO2, and O2 are P(SO3), P(SO2), and P(O2) respectively.
According to the balanced equation, the stoichiometry of the reaction is 2:2:1. So the equilibrium partial pressures can be represented as follows:
P(SO3) = x atm (initial pressure of SO3 - change in pressure)
P(SO2) = 2x atm (initial pressure of SO2 formed is 2x)
P(O2) = x atm (initial pressure of O2 formed is x)
The total pressure at equilibrium is given as 1.017 atm. Therefore:
P(SO3) + P(SO2) + P(O2) = 1.017 atm
Substituting the values:
x + 2x + x = 1.017
4x = 1.017
x = 0.25425 atm
Now that we have the value of x, we can calculate the equilibrium partial pressures:
P(SO3) = 0.25425 atm
P(SO2) = 2 * 0.25425 atm = 0.5085 atm
P(O2) = 0.25425 atm
Finally, we can calculate Kp using the formula:
Kp = (P(SO2))^2 * P(O2) / (P(SO3))^2
Substituting the values:
Kp = (0.5085)^2 * 0.25425 / (0.25425)^2
Kp = 0.129625 / 0.064642
Kp ≈ 2.001
Therefore, the value of Kp for the reaction 2 SO3(g) ⇌ 2 SO2(g) + O2(g) is approximately 2.001.
To find the value of Kp for this reaction, we need to use the equilibrium partial pressures of each gas.
The balanced chemical equation for the reaction is:
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
Let's assume that the partial pressure of SO3, SO2, and O2 at equilibrium is P(SO3), P(SO2), and P(O2), respectively.
According to the ideal gas law, the total pressure (P_total) at equilibrium is the sum of the partial pressures of each gas:
P_total = P(SO3) + P(SO2) + P(O2)
Given that the total pressure (P_total) at equilibrium is 1.017 atm, and the initial pressure of SO3 is 1.000 atm, we can set up the equation:
1.017 atm = P(SO3) + P(SO2) + P(O2)
Since the stoichiometric coefficient for SO3 is 2, the partial pressure of SO2 and O2 will also be 2 times the partial pressure of SO3:
P(SO2) = 2 * P(SO3)
P(O2) = 2 * P(SO3)
Substituting these expressions into the equation above, we get:
1.017 atm = P(SO3) + 2 * P(SO3) + 2 * P(SO3)
1.017 atm = 5 * P(SO3)
Now we can solve for P(SO3):
P(SO3) = 1.017 atm / 5 = 0.2034 atm
Since the stoichiometric coefficient of SO3 is 2, the equilibrium partial pressure of SO3 is half of its initial pressure:
P(SO3 equilibrium) = 0.500 * P(SO3 initial) = 0.500 * 1.000 atm = 0.500 atm
Therefore, the equilibrium partial pressures are:
P(SO3) = 0.2034 atm
P(SO2) = 2 * 0.2034 atm = 0.4068 atm
P(O2) = 2 * 0.2034 atm = 0.4068 atm
Now we can calculate Kp using the expression:
Kp = (P(SO2)^2 * P(O2)) / (P(SO3)^2)
Substituting the values we obtained:
Kp = (0.4068 atm)^2 * (0.4068 atm) / (0.2034 atm)^2
Kp = 0.3342
Therefore, the value of Kp for the given reaction is approximately 0.3342.