A mixture of 0.100 mol NO, 0.200 mol H2 and 0.0800 mol N2 were placed in a 2.00L reaction vessel, heated, and allowed to come to equilibrium conditions. At equilibrium, the molar concentration of N2 was 0.0500 mol/L. Calculate Kc for this reaction.

2NO + 2H2 <---> N2 + 2H2O

I HAVE begun this question by doing the ICE TABLE. but the problem is that at equilibrium, for No it is 0.0500! And the initial is 0.05 so that means it had 0 change in concentration??? And then 2H2O would be 0M at equilibrium. Is that correct? I know how to solve Kc its at the equilibrium concentrations I'm stuck. Thanks.

1 answer

I assume all are gases. I assume initial H2O is 0.
Initial (NO) = 0.100/2 = 0.050M
Initial (H2) = 0.200/2 = 0.100M
Initial (N2) = 0.0800/2 = 0.0400M

........2NO + 2H2 ==> N2 + 2H2O
I......0.05...0.10....0.04..0
C......-2x....-2x....+x....+2x
E...0.05-2x.0.10-2x.0.04+x..2x

At equilibrium (N2) = 0.05; therefore, x ust be 0.01 since 0.04+x = 0.05
Then NO = 0.05-2x
(H2) = 0.10-2x
Kc goes from there.
I think you confused NO with N2.