A 2.00-L vessel contains 1 mol #N_2#, 1 mol #H_2+# and 2 mol #NH_3+#. What is the direction of reaction (forward or reverse) needed to attain equilibrium at #400^oC#? The equilibrium constant #K_c# for the reaction is 0.51 at #400^oC# (rxn below)

Question

#N_2(g)# + #3H_2(g)# #rightleftharpoons# #2NH_3(g)#

Thank you so much in advance!

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Answer 1 · 2018-03-05T09:54:16

You gots #N_2(g)+3H_2(g) rightleftharpoons2NH_3(g)#

#K_c=([NH_3(g)]^2)/([N_2][H_2]^3)=0.51#

Now #Q_"the reaction quotient"=(1.0*mol*L^-1)^2/((0.5*mol*L^-1)(0.5*mol*L^-1)^3)=16#

Since #Q>K_c#...the equilibrium should move to the left as we face the page....