You need to find the arrow keys on your computer and use them.
(N2O) = 0.174mol/4.00L = estimated 0.04
(NO2) = 0.337 mol/4.00L = estd 0.08
...........N2O(g) + NO2(g)==> 3NO(g)
I.........0.04.....0.08........0
C.........-x........-x.........3x
E.........0.04-x...0.08-x......3x
Substitute the E line into Kc expression and solve for x.
Then (NO) = 3x
At 200 °C, Kc = 1.4 x 10-10 for the reaction
N2O(g) + NO2(g) 3NO(g)
If 0.174 mol of N2O and 0.337 mol NO2 are placed in a 4.00 L container, what would the NO concentration be if this equilibrium were established?
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