(NOBr) = 0.782 mols/100L = approx 0.008 but you need to do it more accurately as well as all of the othr values I've estimated.
.......NOBr(g) ↔ NO(g) + 0.5Br2(g)
I......0.008......0..........0
C......-x.........x.......0.5x
E.....0.008-x.....x.......0.5x
Substitute the E line into the Kc expression and solve for Kc. You know 0.5x is 1.42E-3M
A 100 L reaction container is charged with 0.782 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction:
NOBr(g) ↔ NO(g) + 0.5Br2(g)
At equilibrium the bromine concentration is 1.42x10-3 M. Calculate Kc (in M^0.5)
**Not specifying the temperature allows for a more liberal use of random numbers
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