# Chemical Equilibrium

1) Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas

PCl3 + Cl2 ......> PCl5

A gas vessel is charged with a mixture of PCl3 (g) and Cl2 (g) , which is allowed to equilibrate at 450 K . At equilibrium the partial pressures of the three gasses are Ppcl3 =0.124 atm Pcl2 atm = 0.157 atm and Ppcl5 = 1.30 atm

a) what is the value of Kp at this tempreture

b) Does the equilibrium favor reactants or products

I Know the first required , I get Kp= 66.77
but the second I didn't

2)

a) At 800K the equilibrium constant for

I2 ......> 2I

is Kc = 3.1 * 10 ^-5 if an equilibrium mixture in a 10 L vessel contains 2.67 * 10^-2 g of I
how many grams of I2 are in the mixture

b) For 2 SO2(g) + O2(g) ..........> 2SO3(g)

Kp= 3*10^4 at 700 K . In a 2 L vessel the equilibrium mixture contains 1.57g of SO3 and 0.125 g of O2 . How many grams of SO2 are in the vessel ?

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 Apr 27, 2009, 01:24 PM

1. b) Find the total pressure on the LHS and the RHS. The one with the lower pressure will be favoured.

Sorry, I lack of time...

EDIT: oops, sorry, don't consider my answer
 Perito Posts: 3,139, Reputation: 781 Ultra Member #3 Apr 27, 2009, 02:10 PM

$PCl_3 + Cl_2 \rightleftharpoons PCl5$

b. The large partial pressure of PCl5 at equilibrium suggests that PCl5 is favored (products). The large equilibrium constant also suggests the same thing.

2)

a) At 800K the equilibrium constant for

$I_2 \rightleftharpoons 2I$

if $Kc = 3.1\, \cdot \, 10^{-5}$ if an equilibrium mixture in a 10 L vessel contains $2.67\, \cdot\, 10^{-2}$ g of I how many grams of $I_2$ are in the mixture?

$Kc = \frac {[\,I]^2}{[I_2]} = 3.1\, \cdot \, 10^{-5}$

Solve for $[I_2]$. You have the weight of I. You can convert that to moles (the usual concentration units are moles / liter when they calculate Kc).

b) $2 SO_2(g) + O_2(g)\, \rightleftharpoons\, 2SO_3(g)$

$Kp =\frac {[SO_2]^2 \cdot [O_2]}{[SO_3]^2}=3\,\cdot\,10^4$

I won't do the rest for you (I'm too lazy.) You'll have to convert the masses of SO3 and O2 to moles and then use the ideal gas law, $PV=nRT$ at T=700 Kelvins to get the volume of each gas. Since it's not otherwise stated, I would use P = 1 atmosphere.

Once you solve the equation for the pressure of SO2, you'll have to convert that back to the number of moles of SO2 and from that you can calculate the mass.
 Asoom Posts: 52, Reputation: 1 Junior Member #4 Apr 28, 2009, 10:50 PM

I like your explaining , so a lot thanks for your effort .

In 2(a) I converted I to moles , but from where can I get its volume ?

In the b I get [2.25] and when I converted to grams I get 144.5 g

Could you please cheak it if it's right or not ?
 Perito Posts: 3,139, Reputation: 781 Ultra Member #5 Apr 29, 2009, 06:08 AM

2(a)... in a 10 L vessel.

It takes the volume of the vessel - 10 L

2(b)
I'm sure you did it right. Just check your answer by plugging back into what you have.

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