At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the following equation: SbCl5(g)<=> SbCl3(g) + Cl2(g)

Question

(a) An 59.8 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 liter container at 182°C.

1. What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?

2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?

b)If the SbCl5 is 29.2 percent decomposed when equilibrium is
established at 182°C, calculate the values for equilibrium constants Kp and Kc, for this decomposition reaction.
Kc=?
Kp=?

c.In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first
placed in an empty 2.00 liter container maintained at a temperature
different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of
moles of SbCl3 to 0.500 mole at equilibrium?

Answer 1

1. What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
moles = grams/molar mass.
Solve for moles.

M = moles/L. Solve for M.

2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?

Use PV = nRT. Don't forget to change T to Kelvin.

b)If the SbCl5 is 29.2 percent decomposed when equilibrium is
established at 182°C, calculate the values for equilibrium constants Kp and Kc, for this decomposition reaction.
Kc=?
Kp=?

M x fraction decomposed = (SbCl3)
M x fraction decomposed = (Cl2)
M x (1.00 - fraction decomposed) = (SbCl5) after decomposition.
Substitute into the Kc expression below:
Kc = (SbCl3)(Cl2)/(SbCl5) = Kc

For Kp, the following:
You know (Cl2), (SbCl3), and (SbCl5) in M.
Multiply each by 15.0 L (M x L = moles) to obtain moles of each, then use PV = nRT to obtain the pressure of each. Substitute into the Kp expression.
Kp = pSbCl2*pCl2/pSbCl5 = ??

c.In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first
placed in an empty 2.00 liter container maintained at a temperature
different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of
moles of SbCl3 to 0.500 mole at equilibrium? 1mole/2L = 0.5 M
.....SbCl5(g) ==>SbCl3(g) + Cl2(g)
.....0...........0.5M.......0
.................-x........ +x
................0.25M.......y
What is y? 0.5-x = 0.25
Solve for x to obtain y.
Post your work on these if you get stuck.