[osmanrulz]

Ok so here is the question?

1. For the reaction N₂ (g)  + 3H₂   <---> 2NH₃ + 92 kJ
at an equilibrium temperature of 1000 ^o C, a 1.00 L flask contains 0.102 mole of ammonia, 1.03 moles of nitrogen, and 1.62 moles of hydrogen. Calculate the equilibrium cosntant at this temperature.

Now in this case I know I am solving for Kc, which is equal to [Products] / [Reactants].

   [N₂]    [3H₂]     [2NH<sub>3/sub]
I    1.03        1.62                      0.102
C          
E    

I'm not sure How to set up the ice table, I mean is the reaction already at equilibrium, so do the Moles of the compounds, are in the E, or in the I, and how do we know if the reactants, or products is set to 0. What really throws me off is the, extra +92 kJ, i've never dealt with adding energy before.  Are the intitial concentrations all 0??</sub>



      

[Astrokel]

The values should be E as they say at an equilibrium temperature, so just work normally. Don't worry about the enthalpy change, perhaps it is just telling you is an endothermic reaction.