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

Question

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

Caleb Adams · Accepted Answer

At 1500K, the gaseous reversible reaction that is being studied is: #2SO_3(g)rightleftharpoons2SO_2(g)+O_2(g)# Here it is also given that #SO_3(g) and SO_2(g)# are introduced at a constant volume of 300 torr and 150 torr respectively. Since the pressure of a gas  is proportional to the number of moles when their volume and temperature are constant. So we can say that the ratio of number of moles of  #SO_3(g) and SO_2(g)# introduced is #300:150=2:1#. Let these are #2x# mol and #x# mol Write the ICE table now. #color(blue)(2SO_3(g)" "" "rightleftharpoons" "2SO_2(g)" "+" "O_2(g))# #color(red)(I)" "2x" "mol" "" "" "" "" "x" mol"" "" "" "" "0" mol"# #color(red)(C)-2alphax" "mol" "" "+2alphax" mol"" "" "" "alphax" mol"# #color(red)(E)" "(1-alpha)2x" "mol" "(1+2alpha)x" mol"" "" "" "alphax" mol"# where #alpha# represents the degree of dissociation at 1500K So at equilibrium total number of moles of component gases in the reaction mixture is #(2-2alpha+1+2alpha+alpha)x=(3+alpha)x# 
It is also given that at equilibrium the pressure of the reacton mixture is #550"torr"#. Now ratio of the total pressure with the initial pressure of # SO_2(g)# should be equal to the ratio of their respective  number of moles. So #(550"tor")/(150"tor")=((3+alpha)x)/x# #=>alpha+3=11/3#
#=>alpha=11/3-3=2/3# Next, determine the component gases' mole fraction at equilibrium. #chi_(SO_3(g))=((1-alpha)2x)/((3+alpha)x)=((1-2/3)2)/((3+2/3))=2/11# #chi_(SO_2(g))=((1+2alpha)x)/((3+alpha)x)=(1+4/3)/((3+2/3))=7/11# #chi_(O_2(g))=(alphax)/((3+alpha)x)=(2/3)/((3+2/3))=2/11# The partial pressures of the component gases will be if P is the reaction mixture's total pressure at equilibrium. #p_(SO_3(g))=chi_(SO_3(g))xxP=(2P)/11# #p_(SO_2(g))=chi_(SO_2(g))xxP=(7P)/11# #p_(O_2(g))=chi_(O_2(g))xxP=(2P)/11# Now calculation of #color(red)(K_p)# #K_p=(p_(SO_2(g))^2xxp_(O_2(g)))/(p_(SO_3(g))^2)=(((7P)/11)^2xx(2P)/11)/((2P)/11)^2# #=>K_p=(49P)/22# But given value of #P=550" torr"=550/760atm=55/76atm# So #=>K_p=(49xx55)/(22xx76)~~1.61atm# Now calculation of #color(blue)(K_c)# We are aware of the relationship. #color(green)(K_p=K_c(RT)^(Deltan))# where #Deltan="total number of moles  of product gases"-"total number of moles  of reactant gases"# #=>Deltan=(2+1)-2=1# So #K_c=K_p/(RT)# Here #R=0.082LatmK^-1mol^-1# And #T=1500K# By adding these values, we obtain #color(blue)(K_c)=1.61/(0.082xx1500)=1.31xx10^-2#

Naomi Aronson · Answer

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 #"550 torr"#, you can use Dalton's law of partial pressures: #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 #"SO"_2# and #"SO"_3#. if you don't get the right #K_P#, it may also be because your stoichiometric coefficients weren't incorporated into the #K_P# expression. The #color(blue)(K_P)# is then: #K_P = (P_(SO_2)^2P_(O_2))/(P_(SO_3)^2)# #=(("350 torr")^2("100 torr"))/(("100 torr")^2)# #=# #"1225 torr"# Convert to #"atm"# by dividing by #"760 torr/atm"# to get #color(blue)("1.6118 atm")#. Recall that #K_P = K_C*(RT)^(Deltan_"gas")#. Since the mols of gas changed from 2 to 2+1 = 3, we say that #Deltan_"gas" = 1#. Therefore: #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!

Xavier Avery · Answer

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

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