The value of Kp for the following reaction at 627oC is 6.55. SO2(g) + 1/2O2(g) <---> SO3(g) What is Kc at this temperature?

In the actual answer, we used #Deltan_"gas" = 1 - (1 + 1/2) = -1/2# and calculated #K_c# by dividing by #(RT)^(-1//2)#. We imply that #K_P# is based on units of #"atm"#. We found that

#K_c = K_P cdot (1/(RT))^(-1//2) = "56.3 M"^(-1//2)#

where #K_c# has units because we are at NONstandard temperature. If we were at #25^@ "C"#, then we would not write the units.

Below we derive a specific version of the following equation...

#barul|stackrel(" ")(" "K_P = K_c(RT)^(Deltan_"gas")," " Deltan_"gas" = sum_P n_(P) - sum_R n_(R)" ")|#

with #P# being products and #R# being reactants.

Now you don't have to derive this equation for an exam per se... but you do have to know how it works.

Well, we know that for

#1/2"O"_2(g) + "SO"_2(g) -> "SO"_3(g)#

we have that

#K_P = (P_(SO_3))/(P_(SO_2)P_(O_2)^(1//2))#,

and that

#K_c = ([SO_3])/([SO_2][O_2]^(1//2))#

where #[" "]_i# are molar concentrations in #"mol/L"# and the #P_i# are partial pressures in #"atm"#.

Since these are gases, which are said to (pretty much) fully occupy the container space, we could rewrite this as:

#K_c = (n_[SO_3]//V)/((n_[SO_2]//V)(n_[O_2]//V)^(1//2))#

where #V# is the volume of the container and #n# is the mols of gas.

Assuming these are ideal gases,

#PV = nRT => n/V = P/(RT)#,

where #R = "0.082057 L"cdot"atm/mol"cdot"K"#.

Therefore,

#K_c = (P_(SO_3)//RT)/((P_[SO_2]//RT)(P_[O_2]//RT)^(1//2))#

#= (P_(SO_3))/(P_[SO_2]P_[O_2]^(1//2)) cdot (1/(RT))^(1 - (1 + 1/2))#

#= (P_(SO_3))/(P_[SO_2]P_[O_2]^(1//2)) cdot (1/(RT))^(-1//2)#

But the first half of #K_c# as shown here is just #K_P#. Therefore, for this reaction,

#K_c = K_P (1/(RT))^(-1//2)#

Here the #-1/2# reflects the fact that #1/2# a mol of ideal gas was lost going from reactants to products.

Since we are at nonstandard temperature, #K_P# and #K_c# have units.

#K_P = 6.55 (cancel"atm")/(cancel"atm" cdot "atm"^(1//2)) = "6.55 atm"^(-1//2)#

As a result,

#color(blue)(K_c) = K_P (1/(RT))^(-1//2) = K_P (RT)^(1//2)#

#= 6.55 cancel("atm"^(-1//2)) cdot ("0.082057 L"cdotcancel"atm""/mol"cdotcancel"K" cdot (627+273.15 cancel"K"))^(+1//2)#

#= color(blue)("56.3 L"^(1//2)cdot"mol"^(-1//2) = "56.3 M"^(-1//2))#