Asked by Marie Genvieve
At a particular temperature, 12.0 mol of NH3 is introduced into a 2.0 container, and the NH3 partially dissociates by the reaction:
2NH3 <> N2 + 3H2
At equilibrium, 6.0 mol of H2 gas are present. Calculate the K for this reaction.
2NH3 <> N2 + 3H2
At equilibrium, 6.0 mol of H2 gas are present. Calculate the K for this reaction.
Answers
Answered by
DrBob222
I assume that is a 2.0L container but you omitted the unit.
M NH3 = mols/L = 12.0/2.0 = 6.0M
........2NH3 ==> N2 + 3H2
I......6.0.......0.....0
C......-2x.......x.....3x
E......6.0-2x....x.....3x
Kc = (N2)(H2)^3/(NH3)^2
The problems tells you H2 at equil is 6.0mols; therefore, M H2 = 6.0/2.0 = 3.0 M. That means 3x = 3.0M or x = 1.0M
Substitute into Kc expression and solve for Kc.
M NH3 = mols/L = 12.0/2.0 = 6.0M
........2NH3 ==> N2 + 3H2
I......6.0.......0.....0
C......-2x.......x.....3x
E......6.0-2x....x.....3x
Kc = (N2)(H2)^3/(NH3)^2
The problems tells you H2 at equil is 6.0mols; therefore, M H2 = 6.0/2.0 = 3.0 M. That means 3x = 3.0M or x = 1.0M
Substitute into Kc expression and solve for Kc.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.