For the reaction #"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#, at what temperature does #K_c# numerically equal #K_p#?

I got #"12.19 K"#.

For ideal gases, #PV = nRT#. Note that:

• #n/V# is concentration in #"mol/L"#
• #P_("N"_2"O"_4) = (nRT)/(V) = ["N"_2"O"_4]RT#, the partial pressure of #"N"_2"O"_4#.
• #P_("NO"_2) = (nRT)/(V) = ["NO"_2]RT#, the partial pressure of #"NO"_2#.

Recall that stoichiometric coefficients become exponents in equilibrium expressions for gases and aqueous solutions.

So, to relate #K_c# and #K_p#, the expression would be:

#K_c = (["NO"_2]^2)/(["N"_2"O"_4])#

#bb(K_p) = (P_("NO"_2)^2)/(P_("N"_2"O"_4))#

#= (["NO"_2]RT)^2/(["N"_2"O"_4]RT)#

#= bb(RT(["NO"_2]^2)/(["N"_2"O"_4]) = K_cRT#, with units of pressure

Since we are wondering when numerically, #K_c = K_p# for this reaction, we use the expression above in bold to solve for #T# in #"K"#.

#=> RTK_c = K_p# in #"atm"#

Thus, #RT# must numerically equal #1# if #K_c# is to be equal to #K_p#.

#=> color(blue)(T = (K_p)/(RK_c))#

It is implied that #K_p# is typically reported in units of #"atm"# (only #K_p# at standard conditions does not have units). Therefore, we would use #R = "0.082057 L"cdot"atm/mol"cdot"K"# (as opposed to #R = "0.083145 L"cdot"bar/mol"cdot"K"#, or #"8.314472 J/mol"cdot"K"#).

So, #color(blue)(T) = 1/0.082057 * 1/1# #"K"# #~~# #color(blue)("12.19 K")#

The relationship between #sf(K_c)# and #sf(K_p)# is :

#sf(K_p=K_c(RT)^(Deltan))#

#sf(Deltan)# is the no. moles product - no. moles reactant

#sf(Deltan=2-1=1)#

The only way that #sf(K_p)# can have the same numerical value as #sf(K_c)# is when #sf((RT)^1=1)#

#:.sf(T=1/R=1/0.082=12.19color(white)(x)K)#

When the question states #sf(K_p=K_c)# it should state the dimensions that are being used. If pressure was in #sf("N/m"^2)# then the value of 8.31 J/K/mol should be used for R which gives a different value for T.

You need to assume that pressure is normalised against 1 atmosphere and concentration is normalised against unit concentration.

A badly worded question.