PCl3(g)+Cl2(g)⇄PCl5(g) Kc=0.11
A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl5(g) in the container is 0.040M.
On the graph above, carefully draw three curves, one for each of the three gases. The curves must show how the concentration of each of the three gases changes as equilibrium is established. Label each curve with the formula of the gas
(Can you just solve the problem so I can input it into the graph)
As a text-based bot, I am unable to draw graphs for you. However, I can help you solve the problem by providing the equilibrium concentrations for each gas.
Given:
Initial concentration of PCl3 (C1) = 0.60 mol / 1.0 L = 0.60 M
Initial concentration of Cl2 (C2) = 0.70 mol / 1.0 L = 0.70 M
At equilibrium:
Concentration of PCl5 (C3) = 0.040 M
We can use the equilibrium constant expression and the provided equilibrium concentrations to determine the equilibrium concentration of the remaining reactants and products.
Kc = [PCl5] / ([PCl3] * [Cl2])
0.11 = 0.040 / (C1 * C2)
0.11 = 0.040 / (0.60 * 0.70)
0.11 = 0.040 / 0.42
0.11 * 0.42 = 0.040
0.0462 = 0.040
Therefore, the equilibrium concentration of PCl3 will be 0.0462 M and the equilibrium concentration of Cl2 will also be 0.0462 M.
Please note that these values are approximate and rounding may be necessary when constructing your graph.
To solve the problem, we can use the given information to determine the initial concentrations of PCl3(g) and Cl2(g) and then calculate the equilibrium concentrations of PCl5(g), PCl3(g), and Cl2(g).
Given:
Initial volume (V) = 1.0 L
Initial moles of PCl3(g) = 0.60 mol
Initial moles of Cl2(g) = 0.70 mol
Equilibrium concentration of PCl5(g) = 0.040 M
To find the initial concentrations, we divide the moles by the volume:
Initial concentration of PCl3(g) = (0.60 mol) / (1.0 L) = 0.60 M
Initial concentration of Cl2(g) = (0.70 mol) / (1.0 L) = 0.70 M
To determine the equilibrium concentrations, we can use the formula:
Kc = [PCl5(g)] / ([PCl3(g)] * [Cl2(g)])
We'll rearrange the formula to solve for [PCl3(g)] and [Cl2(g)]:
Kc = [PCl5(g)] / ([PCl3(g)] * [Cl2(g)])
Kc * [PCl3(g)] * [Cl2(g)] = [PCl5(g)]
[PCl3(g)] * [Cl2(g)] = [PCl5(g)] / Kc
Plugging in the values:
[PCl3(g)] * [Cl2(g)] = (0.040 M) / (0.11)
Now we can solve for [PCl3(g)] and [Cl2(g)] using the equation:
[PCl3(g)] * [Cl2(g)] = 0.364 M
Since the total initial volume is 1.0 L, the total final volume at equilibrium will also be 1.0 L.
We can use this information to plot the curves on the graph:
- For PCl3(g), the initial concentration is 0.60 M, and at equilibrium, it decreases to a value less than 0.60 M.
- For Cl2(g), the initial concentration is 0.70 M, and at equilibrium, it decreases to a value less than 0.70 M.
- For PCl5(g), the initial concentration is 0 M, and at equilibrium, it increases to 0.040 M.
Please note that the graph must be accurate to properly represent the concentrations of the gases at equilibrium.
To solve this problem, we need to use the given information and apply the principles of chemical equilibrium. The equilibrium constant expression for the reaction is:
Kc = [PCl5] / ([PCl3] * [Cl2])
Given that Kc is 0.11 and the concentrations at equilibrium are [PCl5] = 0.040 M, we can rearrange the equation to solve for the concentrations of PCl3 and Cl2.
0.11 = 0.040 / ([PCl3] * [Cl2])
Let's denote the concentration of PCl3 as x and the concentration of Cl2 as y. Substituting these values into the equation, we get:
0.11 = 0.040 / (x * y)
Rearranging the equation, we have:
x * y = 0.040 / 0.11
Simplifying the right side of the equation, we get:
x * y = 0.3636
Now, we have an equation with two variables. To find the specific values of x and y, we need one more piece of information. Unfortunately, the problem statement does not provide any additional data. Without that information, we cannot accurately determine the particular concentrations of PCl3 and Cl2 at equilibrium.
However, to create the graph, you can assume different values for x and y as long as they satisfy the given conditions. For example, you could assume [PCl3] = 0.3 M and [Cl2] = 1.212 M (or any other combination that gives a product of 0.3636 when multiplied) to draw the curves for each gas.
There isn't enough information to draw a graph. You have the concentrations at zero time and I can calculate the concentrations at equilibrium. There is no way to calculate the numbers between beginning and end because there is no data for that. To begin, the initial concentrations are 0.60 mol/L for PCl3 and 0.70 mol/L Cl2.
...............PCl3 + Cl2 ==> PCl5
I.............0.60.......0.70..........0
C................-x........-x............+x
E...........0.60-x.....0.70-x........x
Kc = 0.11 = (PCl5)/(PCl3)(Cl2)
0.11 = (x)/(0.60-x)(0.70-x)
Solve for x, then evaluate 0.60-x, and 0.70-x. That will give you the final (equilibrium) concentrations. You have the initial concentrations. I presume you just draw a straight line to connect the two point unless you want to try and fake a curved line but you have no point to support that fake line. Post your work on the solution above if you have trouble. Sorry I waited so long to answer. I didn't see it until just now. Good luck.