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3 years ago

Pleaseeeee guysssss

Chemistry

1 answer:

tino4ka555 [31]3 years ago

3 0

Answer:

no answer of this question

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Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

Zarrin [17]

Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M$

$\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M$

$\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial conc. 0.0163 0.00415 0.00276

At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)

As we are given:

Concentration of $H_2$ at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of $HI$ at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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3 years ago