Question

References Use the References to access important values if needed for this question. The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2H19) H2(g) + 12(9) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.301 M HI 4.04x102 MH, and 4.04x10 - MI;. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.18x102 mol of 12(g) is added to the flask? Submit Answer

Answer 1

Answer:

[H₂] = 0,03288 M

[I₂] = 0,05468 M

[HI] = 0,31604 M

Explanation:

For the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

The equilibirum constant, k, is defined as:

k = [H₂][I₂]/[HI]² where k = 1,80x10⁻²

The equilibirum mixture of the system is:

[H₂] = 4,04x10⁻² M

[I₂] = 4,04x10⁻² M

[HI] = 0,301 M

The addition of 2,18x10⁻² mol of I₂ will produce the next changes:

[H₂] = 4,04x10⁻² M - x

[I₂] = 4,04x10⁻² + 2,18x10⁻² - x = 6,22x10⁻² - x

[HI] = 0,301 M + 2x

Because by LeChatelier's principle, the addition of a substance in a chemical equilibrium will produce the increase of the concentration in the other side of the reaction.

Thus, you will obtain:

1,80x10⁻² =  $\frac{[4,04x10^{-2}-x][6,22x10^{-2}-x]}{[0,301 + 2x]^2}$

0,928 x²- 0,124272 x + 8,82062x10⁻⁴

Solving:

x = 0,12639 No chemical sense

x = 0,00752 Real answer.

Thus, final concentrations are:

[H₂] = 4,04x10⁻² M - 0,00752 = 0,03288 M

[I₂] =  6,22x10⁻² - 0,00752 = 0,05468 M

[HI] = 0,301 M + 2×0,00752 = 0,31604 M

I hope it helps!