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...
No comments:
Post a Comment