0.80mols NF3(g) and 0.60mols O2(g) are combined in a 1.00L vessel and allowed to reach equilibrium. At equilibrium, the concentration of NF3(g) is found to be 0.30M. What is the value of the equilibrium constant for this reaction?
Write a balanced chemical equation:
2 NF3(g) + O2(g) ↔ 2 NO(g) + 3 F2(g)
Set up a table to organize the data from the problem:
2 NF3(g) + | O2(g) ↔ | 2 NO(g) + | 3 F2(g) | |
[ ]initial | 0.80 M | 0.60 M | 0 M | 0 M |
Δ[ ] | - 2x | - x | + 2x M | + 3x M |
[ ]equil | (0.80-2x) M | (0.60-x) M | 2x M | 3x M |
Write an expression for the equilibrium constant:
Plug in values from the table:
Determine "x" from the problem. We are told that [NF3]eq=0.30M, and we see in the table that [NF3]eq=(0.80-2x), so:
0.30 = 0.80-2x
x = 0.25
Plugging in:K = 3.4
This equilibrium is (very slightly) product-favored.Practice, practice, practice...
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