Phosphorus pentachloride decomposes according to the chemical equation
PCl5(g) -----> PCl3 (g)+ Cl2(g)
Kc= 1.80 at 250 degrees C
A 0.206 mol sample of PCl5(g) is injected into an empty 3.30 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
I don't know what to do after I set up the Kc equation. How do you isolate x from 1.80=(x^2)/(.062-x)
First, I would check that 0.206/3.30. You shouldn't drop that last number; you are allowed three significant figures in 0.206 and 3.30 and that means you can have three s.f. in the answer. I have 0.0624. Second, your chemistry is right; you've done the hard part. Now it's just algebra. Here is what you start with.
(x^2)
--------- = 1.80
0.0624-x
Multiply both sides by 0.0624-x like this.
(0.0624-x)(x^2)
---------------- = 1.80*(0.0624-x)
(0.0624-x)
You see (0.0624-x) term on the left cancels since it is in the numerator and denominator and you are left with
x^2 = 1.80(0.0624-x)
Note: You could have done the same thing by cross multiplying but I don't know that you've seen that way.
Now multiply the right side to remove the parentheses, move all the terms to the left and you are left with a quadratic which you can solve for x. Let me know if you get stuck. Thanks for showing your work.
I was able to figure it out! Thank you very much for your help(:
To calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium, we can use the equilibrium constant expression and the given information.
Let's assign variables to the concentrations:
Let [PCl5] represent the equilibrium concentration of PCl5(g),
Let [PCl3] represent the equilibrium concentration of PCl3(g),
And let [Cl2] represent the equilibrium concentration of Cl2(g).
According to the balanced chemical equation:
PCl5(g) -----> PCl3(g) + Cl2(g)
The stoichiometry is 1:1:1, meaning:
For every 1 mole of PCl5 that decomposes, 1 mole of PCl3 and 1 mole of Cl2 are formed.
Given that the initial amount of PCl5(g) is 0.206 mol and the reaction vessel has a volume of 3.30 L, we can determine the initial concentration of PCl5(g):
[PCl5] = (0.206 mol) / (3.30 L) = 0.0624 M
Let's assume the change in concentration of PCl5(g) at equilibrium is -x, as it is being consumed. The change in concentration of PCl3(g) and Cl2(g) will also be +x, as they are being formed.
At equilibrium, the concentrations will be:
[PCl5] = 0.0624 - x (M)
[PCl3] = x (M)
[Cl2] = x (M)
We can now write the expression for the equilibrium constant (Kc) in terms of concentrations:
Kc = ([PCl3] * [Cl2]) / [PCl5]
Substituting the known values:
1.80 = (x * x) / (0.0624 - x)
Simplifying the equation:
1.80 * (0.0624 - x) = x^2
Expanding:
0.11232 - 1.80x = x^2
Rearranging the equation to set it equal to zero:
x^2 + 1.80x - 0.11232 = 0
Now, we can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 1, b = 1.80, c = -0.11232
Calculating the discriminant:
D = b^2 - 4ac
D = (1.80)^2 - 4(1)(-0.11232)
D = 3.24 + 0.44864
D = 3.68864
Since the discriminant is positive, we have two solutions for x.
Using the quadratic formula:
x = (-1.80 ± √(3.68864)) / 2
Calculating x:
x₁ = (-1.80 + √(3.68864)) / 2
x₁ = (-1.80 + 1.92125) / 2
x₁ = 0.12125 / 2
x₁ = 0.060625
x₂ = (-1.80 - √(3.68864)) / 2
x₂ = (-1.80 - 1.92125) / 2
x₂ = -3.72125 / 2
x₂ = -1.860625
Since negative concentrations are not physically meaningful, we can discard the negative value. So, x = 0.060625 M.
Now, we can calculate the equilibrium concentrations:
[PCl5] = 0.0624 - x
[PCl5] = 0.0624 - 0.060625
[PCl5] = 0.001775 M
[PCl3] = x
[PCl3] = 0.060625 M
[Cl2] = x
[Cl2] = 0.060625 M
Therefore, at equilibrium:
[PCl5] = 0.001775 M
[PCl3] = 0.060625 M
[Cl2] = 0.060625 M