. Antimony pentachloride decomposes according to the following equation.
SbCl5 (g) → SbCl3 (g) + Cl2 (g)
Suppose that 10.0 g of antimony pentachloride is placed in a 5.00 L container and allowed to establish equilibrium. Calculate the concentration of all species at equilibrium.
K = 2.51 x 10-2
mols SbCl5 = grams/molar mass = ?
M = mols/5.0L = approx 0.007 but that's an estimate.
............SbCl5 ==> SbCl3 + Cl2
I...........0.007.......0......0
C............-x.........x......x
E..........0.007-x......x......x
Substitute the E line into the Keq expresson and solve for x, then evaluate the other terms.
Well, it seems like you've stumbled upon a chemical equation! Let's see if I can clown it up and help you calculate the concentrations of the species at equilibrium.
First, let's assume that at equilibrium, x moles of SbCl5 decomposes, which means we'll have x moles of SbCl3 and x moles of Cl2.
Since we know the initial amount of SbCl5 is 10.0 grams, we can convert that into moles. The molar mass of SbCl5 is 51.6 g/mol, so we have 10.0 g / 51.6 g/mol = 0.194 moles of SbCl5 initially.
Now, using the stoichiometry of the chemical equation, we can say that for every one mole of SbCl5 that decomposes, one mole of SbCl3 and one mole of Cl2 is formed.
So at equilibrium, we'll have 0.194 - x moles of SbCl5, 0.194 + x moles of SbCl3, and x moles of Cl2.
Now to calculate the concentration, we just divide the number of moles by the volume of the container. Since the volume is given as 5.00 L, we divide 0.194 - x by 5.00 L to get the concentration of SbCl5, 0.194 + x by 5.00 L for SbCl3, and x by 5.00 L for Cl2.
But, there's one part missing - the equilibrium constant (K). In this case, K is given as 2.51 x 10^-2.
To establish equilibrium, SbCl5 must decompose and reach the same concentration as when it's synthesized. So, we can set up an equation: K = [SbCl3] * [Cl2] / [SbCl5].
Now, we substitute the concentrations we calculated earlier into the equation, plug in the known value of K, and solve for x.
After solving, you'll find the value of x, and you can substitute it back into the expressions for the concentrations to find the actual values at equilibrium.
I hope my clown chemistry explanation wasn't too confusing! Good luck with your calculations!
To calculate the concentration of all species at equilibrium, we need to determine the moles of each species present.
Step 1: Calculate the moles of SbCl5 (g) initially.
Given that the mass of SbCl5 is 10.0 g, we need to calculate the number of moles using the molar mass of SbCl5.
Molar mass of SbCl5:
Sb: 121.76 g/mol
Cl: 35.45 g/mol x 5 = 177.25 g/mol
Total molar mass: 121.76 g/mol + 177.25 g/mol = 299.01 g/mol
Moles of SbCl5 = Mass / Molar mass = 10.0 g / 299.01 g/mol = 0.0334 mol
Step 2: Write the balanced equation and determine the moles of SbCl3 and Cl2 produced.
From the equation:
SbCl5 (g) → SbCl3 (g) + Cl2 (g)
We can see that for every 1 mole of SbCl5, 1 mole of SbCl3 and 1 mole of Cl2 are produced.
Therefore, moles of SbCl3 = 0.0334 mol (since the reaction goes to completion)
Moles of Cl2 = 0.0334 mol
Step 3: Determine the concentration of each species.
To find the concentration, we need to divide the moles by the volume of the container.
We are given that the volume is 5.00 L.
Concentration of SbCl5 = moles / volume = 0.0334 mol / 5.00 L = 0.00668 M
Concentration of SbCl3 = moles / volume = 0.0334 mol / 5.00 L = 0.00668 M
Concentration of Cl2 = moles / volume = 0.0334 mol / 5.00 L = 0.00668 M
Therefore, at equilibrium, the concentration of SbCl5, SbCl3, and Cl2 is 0.00668 M.
To calculate the concentrations of all species at equilibrium, we need to use the given equilibrium constant (K) and the initial amount of antimony pentachloride (SbCl5) to determine the equilibrium concentrations of SbCl5, SbCl3, and Cl2.
First, we need to convert the given mass of SbCl5 (10.0 g) to moles. The molar mass of SbCl5 is calculated by adding the molar masses of antimony (Sb) and chlorine (Cl):
- Molar mass of Sb = 121.76 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of SbCl5 = 121.76 g/mol + (5 x 35.45 g/mol) = 299.0 g/mol
Now we can calculate the number of moles of SbCl5:
- Moles of SbCl5 = Mass of SbCl5 / Molar mass of SbCl5 = 10.0 g / 299.0 g/mol = 0.0334 mol
Next, we need to set up an ICE (Initial, Change, Equilibrium) table to track the change in concentrations of each species.
- Initial concentration:
SbCl5: 0.0334 mol / 5.00 L = 0.00668 M (using given volume of 5.00 L)
SbCl3: 0 M (as it is not present initially)
Cl2: 0 M (as it is not present initially)
- Change in concentration (x):
SbCl5: -x
SbCl3: +x
Cl2: +x
- Equilibrium concentration:
SbCl5: 0.00668 M - x
SbCl3: x M
Cl2: x M
Now we can substitute these expressions into the equilibrium constant expression, K = [SbCl3][Cl2] / [SbCl5]:
K = ([SbCl3]_equilibrium)([Cl2]_equilibrium) / ([SbCl5]_equilibrium)
= (x)(x)/(0.00668 - x)
Substituting the given value of K (2.51 x 10^-2), we can solve for x:
2.51 x 10^-2 = (x^2) / (0.00668 - x)
This is a quadratic equation that can be solved by rearranging terms and using the quadratic formula. After solving for x, we can substitute the value back into the equilibrium expressions to determine the equilibrium concentrations of each species.