The equilibrium constant (Kp) for the interconversion of PCl5 and PCl3 is 0.0121:
PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
A vessel is charged with PCl5, giving an initial pressure of 0.123 atm. At equilibrium, the partial pressure of PCl3 is __________ atm.
I have a hit a wall with this question, this is where I got to:
PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
I: 0.123 0 0
C: -2x. x x
E: 0.123-2x x x
Kp= [PCl3][Cl2]/[PCl5]
Kp=[x][x]/[0.123-2x]
0.0121=x^2/0.123
and thats where I have gotten stuck can't seem to get the correct answer, I know is 0.0330 but I can not figure out how to exactly get to it. Thanks for any help.
First, that C line is -x.........x.......x and not -2x
ane the E line is 0.123-x.......x.......x
Then you have Kp = 0.0121 = (x)(x)/(0.123-x) and solve that quadratic. I get 0.032996 which rounds to 0.0330
Well, it seems like you're in a bit of a pickle with this question. But don't worry, I'm here to help and hopefully bring a smile to your face!
Let's take a closer look at where you left off. You correctly set up the expression for the equilibrium constant (Kp) using the concentrations of the products and reactant:
Kp = [PCl3][Cl2] / [PCl5]
And then you substituted the concentrations in terms of 'x':
0.0121 = (x^2) / (0.123 - 2x)
Now, to solve for 'x', we can rearrange the equation a bit:
0.0121(0.123 - 2x) = x^2
0.0015 - 0.0242x + 0.0242x^2 = x^2
0.0242x^2 - x^2 + 0.0242x - 0.0015 = 0
Combine like terms:
0.0232x^2 + 0.0242x - 0.0015 = 0
Now, we can use the quadratic formula to solve for 'x':
x = (-0.0242 ± sqrt((0.0242)^2 - 4(0.0232)(-0.0015))) / (2(0.0232))
Now, when you plug in the values and crunch the numbers, you'll get two possible values for 'x'. However, we're dealing with partial pressures, so the negative value for 'x' doesn't make sense in this context. We can discard the negative value.
So, the final value for 'x' would be the positive root, which gives you the partial pressure of PCl3 at equilibrium:
x = 0.0330 atm
And that's how you arrive at the answer of 0.0330 atm. I hope this explanation brought a little bit of laughter to your studying session! If you have any more questions, I'm here to help, just like a clown with an endless supply of balloons!
To find the partial pressure of PCl3 at equilibrium, we can set up an equation using the given equilibrium constant (Kp).
The equilibrium constant expression for the given reaction is:
Kp = [PCl3] * [Cl2] / [PCl5]
We know that the initial pressure of PCl5 is 0.123 atm. At equilibrium, let's assume that the pressure of PCl3 is "x" atm, and the pressure of Cl2 is also "x" atm.
Substituting these values into the equilibrium constant expression:
0.0121 = (x * x) / (0.123 - 2x)
Now, let's solve this equation to find the value of "x":
0.0121 * (0.123 - 2x) = x^2
0.00149 - 0.0242x = x^2
Rearranging the equation:
x^2 + 0.0242x - 0.00149 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-0.0242 +/- sqrt((0.0242)^2 - 4 * 1 * (-0.00149))) / (2 * 1)
x = (-0.0242 +/- sqrt(0.00058564 + 0.00596)) / 2
x = (-0.0242 +/- sqrt(0.00654564)) / 2
x = (-0.0242 +/- 0.08089) / 2
Taking both positive and negative values:
x1 = (-0.0242 + 0.08089) / 2 = 0.028345
x2 = (-0.0242 - 0.08089) / 2 = -0.052545
Since the values of partial pressures cannot be negative, we discard x2 (-0.052545).
Therefore, the partial pressure of PCl3 at equilibrium is approximately 0.0283 atm (rounded to 4 decimal places).
Note: It's important to remember that the quadratic equation may have two roots, but in this case, one of them is not physically possible, so we only consider the positive root.
To solve this problem and find the partial pressure of PCl3 at equilibrium, you can use the expression for the equilibrium constant (Kp) and the given initial pressure of PCl5.
First, you correctly set up the equilibrium expression as:
Kp = [PCl3][Cl2] / [PCl5]
Let's substitute the given values into the expression:
0.0121 = (x)(x) / (0.123 - 2x)
Now, let's solve for x. Start by multiplying both sides of the equation by (0.123 - 2x):
0.0121 * (0.123 - 2x) = x^2
0.0014923 - 0.0242x = x^2
Rearrange the equation to form a quadratic equation:
x^2 + 0.0242x - 0.0014923 = 0
Now, you can solve this quadratic equation for x. Since the equilibrium pressure should be positive, we'll pick the positive root.
Using a quadratic formula or a calculator, the positive root of this equation is approximately x ≈ 0.0330.
Therefore, the partial pressure of PCl3 at equilibrium is approximately 0.0330 atm.