Consider the following equilibrium:
PCl5 = PCl3+ Cl2
5.0 moles of are placed in a 10.0 L flask at 200 C and allowed to come to equilibrium. Analysis shows that 1.0 mole of is present in the equilibrium mixture. How many moles of are present at equilibrium.
Choose one answer.
a. 0
b. 1.0
c. 2.0
d. 3.0
e. 4.0
I get ) but not sure if I am answering correctly based on what they give in the problem.
I get 0 as the answer but not sure if I am right
You didn't type all of the problem.
Analysis shows that 1 mole of (WHAT) is present at equilibrium.
To solve this problem, you can use the principle of the Law of Mass Action. According to the equation given, PCl5 decomposes into PCl3 and Cl2. Let's assume that x moles of PCl5 decompose to form PCl3 and Cl2 at equilibrium.
The balanced equation for the reaction is:
PCl5 → PCl3 + Cl2
Initially, you have 5.0 moles of PCl5 in a 10.0 L flask, so the initial concentration of PCl5 ([PCl5]initial) is 5.0 mol/10.0 L = 0.5 mol/L.
At equilibrium, the concentration of PCl5 ([PCl5]equilibrium) will be 0.5 mol/L - x mol/L. The molar concentration of PCl3 ([PCl3]equilibrium) and Cl2 ([Cl2]equilibrium) will be x mol/L each.
According to the problem, the analysis shows that 1.0 mole of PCl3 is present at equilibrium.
So, we can set up an equation using the Law of Mass Action:
Kc = ([PCl3]equilibrium * [Cl2]equilibrium) / [PCl5]equilibrium
Kc is the equilibrium constant, which is given by:
Kc = [PCl3]^1 * [Cl2]^1 / [PCl5]^1
Plugging in the concentrations, we have:
Kc = (x * x) / (0.5 - x)
The equilibrium constant for this reaction is a constant at a given temperature. Since the temperature is specified as 200 ˚C, we can assume that Kc is a constant value.
Now, we can solve for x:
Kc = (x * x) / (0.5 - x)
Since we know [PCl3]equilibrium = 1.0 mole and [Cl2]equilibrium = x, we can substitute these values into the equation:
Kc = (1.0 * x) / (0.5 - x)
Simplifying the equation gives:
0.5 - x = x
0.5 = 2x
x = 0.25 mol/L
Therefore, at equilibrium, the concentration of PCl3 is 1.0 mol/L and the concentration of PCl5 is 0.5 mol/L - 0.25 mol/L = 0.25 mol/L.
Since the question asks for the number of moles of PCl5 present at equilibrium and we know the concentration of PCl5 is 0.25 mol/L, we can calculate the number of moles using the formula:
moles = concentration * volume
moles = 0.25 mol/L * 10.0 L = 2.5 moles
So, the correct answer is c. 2.0 moles.
To solve this problem, we can use the principles of stoichiometry and the concept of the equilibrium constant (Kc).
First, let's write the balanced equation for the given equilibrium reaction:
PCl5 ⇌ PCl3 + Cl2
From the equation, we can see that the stoichiometry is 1:1:1. This means that for every mole of PCl5 that reacts, we will form 1 mole each of PCl3 and Cl2.
Given that 1.0 mole of PCl5 is present in the equilibrium mixture, we can conclude that 1.0 mole of PCl3 and 1.0 mole of Cl2 are also present, according to the stoichiometry.
Therefore, the total number of moles of PCl3 and Cl2 in the equilibrium mixture is 1.0 mole + 1.0 mole = 2.0 moles.
So, the correct answer is:
c. 2.0 moles