mols N2 = 2.0/28 = about 0.07
(N2) = 0.07/1.3L = about 0.05
mols O2 = 3.0/32 = about 0.09
(O2) = a0.09/1.L = about 0.07
All of the above need to be recalculated more accurately.
.......N2 + O2 ==> 2NO
I....0.05..0.07......0
C.....-x...-x.......2x
E..0.05-x..0.07-x....2x
Substitute the E line into the Keq expression and solve for x, then evaluate the N2 and O2.
The second part is done the same way but you start with NO as 1.5/1.8 = ? and N2 and O2 are zero. The E line will be x for N2, x for O2, and ?-2x for NO.
Post your work if you get stuck.
At 3745°C, K = 0.093 for the following reaction.
N2(g) + O2(g) equilibrium reaction arrow 2 NO(g)
Calculate the concentrations of all species at equilibrium for each of the following cases.
(a) 2.0 g N2 and 3.0 g O2 are mixed in a 1.3-L flask.
[N2]
[O2]
[NO]
(b) 1.5 mol pure NO is placed in a 1.8-L flask.
[N2]
[O2]
[NO]
1 answer
1 answer