For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5.
PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g)
Suppose that 2.030 g of PCl5 is placed in an evacuated 535 mL bulb, which is then heated to 600. K.
(a) What would be the pressure of PCl5 if it did not dissociate?
(b) What is the partial pressure of PCl5 at equilibrium?
(c) What is the total pressure in the bulb at equilibrium?
(d) What is the degree of dissociation of PCl5 at equilibrium?
PLEASE HELP
a. USE PV = nRT. Let's just say P is 1 atm. It isn't but let's go with that.
b. PCl5 ==> PCl3 + Cl2
I.......1 atm.......0........0
C.......-p...........p.........p
E.....1-p...........p..........p
Then Kp = 11.5 = (p)(p)/(1-p)
Solve for p and evalute 1-p
c. Ptotal = p(PCl5) + p(PCl3) + p(Cl2) = ?
d. degree of dissociation = [p(PCl3)/p(PCl5)]
Post your work if you get stuck.
NOTE: Answers given by HONEY. Check that 1.263 carefully. I obtained a different answer.
Also the ICE chart is not set up properly; i.e., equilibrium p for PCl5 is 1.263-y but the 1.263 value is in question.
For d the degree of dissociation is a typo; i.e., the first PCl5 should be PCl3
My apologies, thanks for clarifying
Sure, I'd be happy to help! But before we start, I have a joke for you: Why don't scientists trust atoms? Because they make up everything! Now, let's get to the questions:
(a) If PCl5 did not dissociate, the pressure of PCl5 would be the initial pressure of PCl5. However, we are not given the initial pressure, so we'll move on to the next question.
(b) To find the partial pressure of PCl5 at equilibrium, we need to use the equilibrium constant, Kp. Since we are given the value of Kp (11.5), we can use the equation: Kp = (PCl3)(Cl2) / (PCl5). We can assume that there is no PCl3 or Cl2 at the start, so:
Kp = (PCl3)(Cl2) / (PCl5) = (0)(0) / (PCl5) = 0 / (PCl5) = 0
Therefore, the partial pressure of PCl5 at equilibrium is zero.
(c) The total pressure in the bulb at equilibrium is the sum of the partial pressures of all the gases present. In this case, we have PCl3 and Cl2 as well. Since we know the partial pressure of PCl5 is zero (from part b), and we are not given any information about the partial pressures of PCl3 and Cl2, we cannot determine the total pressure in the bulb at equilibrium.
(d) The degree of dissociation of PCl5 at equilibrium is a measure of how much PCl5 dissociates into PCl3 and Cl2. It can be calculated using the formula: degree of dissociation = (partial pressure of PCl5 at equilibrium) / (initial pressure of PCl5). However, since we don't know the initial pressure of PCl5, we can't calculate the degree of dissociation.
I hope this helps! Let me know if you have any more questions, or if you need another joke to lighten the mood.
Sure, I can help you with this problem. Let's go step by step.
(a) To determine the pressure of PCl5 if it did not dissociate, we can assume that PCl5 remains as PCl5 without any dissociation. Therefore, the pressure of PCl5 in the bulb would be the same as the initial pressure.
(b) To find the partial pressure of PCl5 at equilibrium, we need to use the equilibrium constant (Kp). The equilibrium constant expression for the given reaction is:
Kp = (PCl3)(Cl2) / (PCl5)
We can rearrange this expression to solve for the partial pressure of PCl5:
PCl5 = (PCl3)(Cl2) / Kp
However, we need the partial pressures of both PCl3 and Cl2 to calculate PCl5. Unfortunately, we don't have that information given in the problem. We can't proceed without knowing the partial pressures of PCl3 and Cl2 when the system reaches equilibrium.
(c) The total pressure in the bulb at equilibrium can be calculated by summing up the partial pressures of all the components (PCl5, PCl3, and Cl2) at equilibrium. However, as mentioned earlier, we need the partial pressures of PCl3 and Cl2 to calculate the total pressure. Therefore, we can't determine the total pressure without additional information.
(d) The degree of dissociation (α) of PCl5 at equilibrium can be calculated using the ideal gas law. The equation to find α is:
α = (n(PCl3) + n(Cl2)) / n(PCl5)
Where n is the number of moles. We need the moles of each gas component to proceed with this calculation. To find the moles, we can use the given mass of PCl5 and the molar mass of PCl5.
To summarize, without the partial pressures of PCl3 and Cl2, we cannot answer part (b) or part (c) of the question. Similarly, without the number of moles of each component, we cannot answer part (d) of the question.
Please double-check the information provided or provide additional details to proceed with solving the problem.
a. 1.263 IF the unit is atmospheres.
Here's the rest of the answers too :)
b. Set up at ICE chart.
Initial: pressure PCl5 = 1.263 atm
partial p PCl3 = 0
partial p Cl2 = 0
change: partial p PCl3 = +y
partial p Cl2 = y
partial pressure PCl5 = -y
equilibrium pressures:
add the columns to obtain
partial p PCl3 = y
partial p Cl2 = y
partial pressure PCl5 = 1.263-y
Now plug all that into the equilibrium constant expression; Kp = ------
and solve for y, the only unknown in the equation. I expect you will get a quadratic equation. That will give you y and 1.263-y (1.263-y will be the partial pressure of PCl5).
c. Add partial pressure PCl5 + partial pressure PCl3 + partial pressure Cl2.
d. The degree of dissociation is
partial p PCl5 at equilibrium/partial p PCl5 initially. By the way, if you then multiply that by 100 you will have percent dissociation. The degree of dissociation IS NOT expressed in percent terms.