Chemistry, 29.11.2019 07:31 heathkid23

$h_2(g) + i_2(g) \longrightarrow 2hi(g)$
the forward reaction above is exothermic. at equilibrium, what happens if the reaction mixture is cooled at constant volume?
select all that apply.
1. the reaction absorbs energy.
2. the reaction releases energy.
3. [h₂] and [i₂] increase.
4. [h₂] and [i₂] decrease.
5. [h₂] and [i₂] remain constant.
6. [hi] increases.
7. [hi] decreases.
8. [hi] remains constant.

if c is added to the equilibrium system above, in which direction will the equilibrium shift?

Answers: 1

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Answer from: heids17043

The appropriate options are:

2. The reaction releases energy.

4. $[H_{2}]$ and $[I_{2}]$ increase.

6. [HI] increases.

The addition of a non-reacting component C will have no effect on the equilibrium system at constant volume.

Explanation:

$H_{2}+I_{2}\longrightarrow2HI(exothermic)$

The question can be answered by using Le Chatelier's principle.

According to Le Chatelier's Principle, for a exothermic reaction, decreasing the temperature at equilibrium will cause the forward reaction to occur.

If the forward reaction occurs, subsequently the concentrations of the reactants will decrease and that of the product will increase.

This is the reason for the options 4 and 6 being correct.

Exothermic reactions are the ones that release energy and endothermic reactions are the ones that absorb energy.

Since the forward reaction is exothermic, cooling the reaction mixture at constant volume, will lead to a release of energy.

This is the reason for the option 2 to be correct.

If a component C is added at constant volume, there will be no effect on the equilibrium system, assuming that the component C is non-reacting.

The concentration of that component C will increase, at constant volume.

Answer from: Quest

It could possibly be beneficial if the experiment is successful.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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[tex]h_2(g) + i_2(g) \longrightarrow 2hi(g)[/tex]
the forward reaction above is exothermic. at...

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