# Question a90d2

Mar 22, 2014

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)# / ${K}_{c}$

${K}_{p}$ = ${\left(0.08206 X 1338\right)}^{- 2}$ / 2.8 x ${10}^{- 9}$
= 1 / ${\left(0.08206 X 1338\right)}^{2}$ x 2.8 x ${10}^{- 9}$

${K}_{p}$ = 1/ 33755 x ${10}^{- 9}$ = 29626.