.........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)
I.........2M........1M..........2M
C........-2x........-3x.........2x
E..........?.......2.53M.........?
Since we know eq (H2) = 2.53, that means that H2 was formed, the reaction is proceeding from right to left and that means we must change the sign of the C line. I will do that here.
........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)
I.......2M........1M..........2M
C......+2x........+3x.........-2x
E.......?.......2.53M..........?
Then we know 1 + 3x = 2.53 which means x must be 0.51
When you know x you can calculate all of the C line. So equilibrium N2 must be 0.51 x 2 and add 2 and equilibrium NH3 must be 0.51 x 2 and that subtracted from 2M. Plug into Keq expression and solve for Kc.
A 1.000 L vessel is filled with 2.000 moles of
N2, 1.000 mole of H2, and 2.000 moles of NH3.
When the reaction
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
comes to equilibrium, it is observed that the
concentration of H2 is 2.53 moles/L. What is
the numerical value of the equilibrium constant
Kc?
1 answer
1 answer