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gogolik [260]
3 years ago
14

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

ner initially holds 0.00623 mol of H2, 0.00414 mol of I2, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H2(g) is 0.00467 M. What are the concentrations of HI(g) and I2(g)?
Chemistry
1 answer:
Zarrin [17]3 years ago
8 0

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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Answer:

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Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

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We were given the following:

Total pressure = 2.45 atm

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3 years ago
½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole 21.0 kcal/mole + C(s) + 2S(s) → CS2(l) What type of reaction is represented by the
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<u>Answer:</u>

Exothermic Reaction are those reaction, in which energy is released while in endothermic reaction are those, in which energy is absorbed.

<u>Explanation:</u>

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so it is <em>exothermic reaction</em>

Second reaction:

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