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davidcstro851
13.09.2019 •
Chemistry
Hydrogen iodide decomposes according to the equation:
2hi(g)h2(g)
+ i2(g),kc = 0.0156 at 400oc
a 0.0660 mole sample of hi was injected into a 2.00 l
reactionvessel held at 400oc.
calculate the concentration of hi and i2
andh2 at equilibrium.
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Ответ:
[HI] = 0,0264 M
[H₂] = 0,003299 M
[I₂] = 0,003299 M
Explanation:
For the equation:
2 HI (g) ⇄ H₂(g) + I₂(g) With k= 0,0156. Also, k=
(1)
The initial concentration of HI is:
[HI] =
= 0,0330 M
The concentration of the reactant and products in equilibrium is:
[HI] = 0,033 M - 2x ⇒ because 2 moles are consumed
[H₂] = x ⇒ because 1 mole is produced
[I₂] = x ⇒ Also, 1 mole is produced
Thus, replacing in (1):
0.0156 =![\frac{[x]^2}{[0,0330-2x]^2}](/tpl/images/0230/3523/4723e.png)
The quadratic equation to solve is:
0,9376x² + 0,0020592x - 0,000017 = 0
Have two solutions:
x = -0,0054955 ⇒ No physical sense. Will produce negative concentrations
x = 0,003299⇒ Real answer
Thus, concentrations in equilibrium are:
[HI] = 0,033 - 2×0,003299 = 0,0264 M
[H₂] = 0,003299 M
[I₂] = 0,003299 M
I hope it helps!
Ответ: