Question

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.

Answer 1

M (HI) = 0.033MHI

M (N2) = 0.0165MN2

M (H2) = 0.0165MH2

Step-by-step explanation:

The Molarity formula is the amount of substance in moles (n) divided by one liter of solution (L)

M = n / L

M (HI) = n / V = ​​0.066mol HI / 2L solution

M (HI) = 0.033MHI

moles of H2 = 0.066molHI * (1mol H2 / 2 molesHI) = 0.033mol H2

moles of

M (H2) = n / V = ​​0.033molH2 / 2Ldisolution

M (H2) = 0.0165MH2

N2 = 0.066molHI * (1mol N2 / 2 molesHI) = 0.033mol N2

M (N2) = n / V = ​​0.033molN2 / 2L solution

M (N2) = 0.0165MN2

Answer 2

[HI] = 0,0264 M

[H₂] = 0,003299 M

[I₂] = 0,003299 M

Step-by-step explanation:

For the equation:

2 HI (g) ⇄ H₂(g) + I₂(g) With k= 0,0156. Also, k=
`([H_(2)[I_(2) ]  )/([HI]^(2))` (1)

The initial concentration of HI is:

[HI] =
`(0,0660)/(2,00)` = 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 =
`([x]^2)/([0,0330-2x]^2)`

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!