A 100 L reaction container is charged with 0.762 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 1.56x10-3 M. Calculate Kc (in M0.5)
**Not specifying the temperature allows for a more liberal use of random numbers.
2 answers
I have no idea what (in M0.5) stands for. M means mols/L but the 0.5 leaves me stumped. Please help; perhaps I can work the problem if I know what that stands for. Thanks.
I have no idea what (in M0.5) stands for. M means mols/L but the 0.5 leaves me stumped. Anyway-------
You have 0.762 mols in 100 L so (NOBr) = 0.762 moles/100 L = 0.00762 M initially.
....................NOBr(g) ↔ NO(g) + 0.5Br2(g)
I.....................0.00762M.......0...........0
C.......................-x................+x.........0.5x
E...............0.00762 - x............x....... 0.5x
but the problem tells you that (Br2) = 0.00156 M = 0.5x
Kc = (NO)(Br2)^0.5/(NOBr)
Evaluate NOBr, NO and Br2 then
Substitute the E line into the Kc expression and solve for Kc. Post your work if you get stuck. I get an answer of 0.027 but it's late late and past my bed time. Be sure you check that out carefully.
You have 0.762 mols in 100 L so (NOBr) = 0.762 moles/100 L = 0.00762 M initially.
....................NOBr(g) ↔ NO(g) + 0.5Br2(g)
I.....................0.00762M.......0...........0
C.......................-x................+x.........0.5x
E...............0.00762 - x............x....... 0.5x
but the problem tells you that (Br2) = 0.00156 M = 0.5x
Kc = (NO)(Br2)^0.5/(NOBr)
Evaluate NOBr, NO and Br2 then
Substitute the E line into the Kc expression and solve for Kc. Post your work if you get stuck. I get an answer of 0.027 but it's late late and past my bed time. Be sure you check that out carefully.
3 answers