# Someone would be kind enough to help me with this exercise: #2"SO"_3(g) -> 2"SO"_2(g) + "O"_2(g)# ?

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Someone would be kind enough to help me with this exercise: At 1500 k they are introduced at a constant volume of 300 torr and 150 torr of SO3, SO2. It establishes the following equilibrium: #2SO_3-> 2SO_2 + O_2# to the equilibrium pressure that is 550 torr. Calculate kp and kc. Answers: Kp: 1.61 atm Kc: 1.31 x 10-2 pls help me fast . please show the working.

how these answers came i dont understand

there can be mistakes in the question . i think it is incomplete . what is incomplete

Someone would be kind enough to help me with this exercise: At 1500 k they are introduced at a constant volume of 300 torr and 150 torr of SO3, SO2. It establishes the following equilibrium:

how these answers came i dont understand

there can be mistakes in the question . i think it is incomplete . what is incomplete

##### 2 Answers

**The gaseous reversible reaction under consideration at 1500K is:**

#2SO_3(g)rightleftharpoons2SO_2(g)+O_2(g)#

Here it is also given that

Now write the **ICE table**

#color(blue)(2SO_3(g)" "" "rightleftharpoons" "2SO_2(g)" "+" "O_2(g))#

where

So **at equilibrium** total number of moles of component gases in the reaction mixture is

It is also given that at equilibrium the pressure of the reacton mixture is

Now ratio of the total pressure with the initial pressure of

So

Now calculating **mole fraction** of the component gases at equilibrium

If P be the total pressure of the reaction mixture at equilibrium the the **partial pressures** of component gases will be

**Now calculation of**

But given value of

So

**Now calculation of**

We know the relation

where

So

Here

And

Inserting these values we get

Here's another way to do it. Your reaction was:

#2"SO"_3(g) rightleftharpoons 2"SO"_2(g) + "O"_2(g)#

Since you have a constant volume, and since the temperature is assumed constant as well (as you are not given two temperatures), you can expect that *the change in mols of gas relates mainly with the change in pressure*, meaning that

#P = P_1 + P_2 + . . . # , Dalton's Law of Partial Pressures,

applies, and **the given equilibrium pressure is the total pressure of all gases in the mixture.**

Filling out an ICE Table gives:

#" "" "" "2"SO"_3(g) rightleftharpoons 2"SO"_2(g)" " + " ""O"_2(g)#

#"I"" "" ""300 torr"" "" ""150 torr"" "" "" "" ""0 torr"#

#"C"" "" ""-2x torr"" "" ""+2x torr"" "" "" ""+x torr"#

#"E"" "" ""300-2x torr"" ""150+2x torr"" "" ""x torr"#

Remember that the change in pressure will include the *stoichiometric coefficients* in front of the molecule in the balanced reaction.

But since you know that the equilibrium pressure was

#P = (300 - 2x) + (150 + 2x) + x = 550#

#P = 450 + x = 550#

#color(green)(x = "100 torr")#

That gives you each equilibrium partial pressure as:

#P_(SO_3) = 300 - 2(100) = "100 torr"#

#P_(SO_2) = 150 + 2(100) = "350 torr"#

#P_(O_2) = "100 torr"#

Note that if you get a negative pressure, it means you mixed up the partial pressures of

The

#K_P = (P_(SO_2)^2P_(O_2))/(P_(SO_3)^2)#

#=(("350 torr")^2("100 torr"))/(("100 torr")^2)#

#=# #"1225 torr"#

Convert to

Recall that

#color(blue)(K_C) = ("1.61 atm")/(("0.082057 L"cdot"atm/mol"cdot"K")("1500 K"))#

#= 0.013095 = color(blue)(1.31 xx 10^(-2) "mol/L")#

though it tends to be reported without units. Hope that helps!