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Answer 1 · 2014-03-22T06:14:57

For the reaction, 2 N#H_3# (g) <-----> #N_2# (g) + 3 #H_2# (g) at 298 K, #K_c# = 2.8 x #10^ (-9)# What is the value #K_p# of for this reaction?

#K_c# = #(RT)^(∆n)#. #K_p#

Change in number of moles : total number of moles of products - total number of moles of reactants.
∆n = ∑ #n_ p# - ∑ #n_ r#

∆n = (1+3) - 2 = -2

#K_c# = #(RT)^(∆n)#. #K_p#

#K_p# = #(RT)^(∆n)# / #K_c#

#K_p# = #(0.08206 X 1338)^(-2)# / 2.8 x #10^ (-9)#
= 1 / #(0.08206 X 1338)^(2)# x 2.8 x #10^ (-9)#

#K_p# = 1/ 33755 x #10^ (-9)# = 29626.