Chemistry, 15.03.2020 05:26 roman2328

Consider the reaction H2(g) + I2(g) <---> HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 525. Which direction will the system shift to?
A. The equilibrium will shift to the left to favor the reactants.
B. The equilibrium will shift to the right to favor the products.
C. The equilibrium will not shift in any direction.
D. The equilibrium will shift to the forward reaction.

Consider the reaction H2(g) + I2(g) <---> HI(g) with an equilibrium constant of 46.3 and a rea

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Answer from: noahmccall647

A. The equilibrium will shift to the left to favor the reactants.

Explanation:

Step 1: Data given

K = 46.3

Q = 525

Step 2: The balanced equation

H2(g) + I2(g) <---> HI(g)

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

For this equation Q >> K

This means there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

A. The equilibrium will shift to the left to favor the reactants.

Answer from: Quest

Answers are 1,2,4 its correct on edgenuity.

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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Consider the reaction H2(g) + I2(g) <---> HI(g) with an equilibrium constant of 46.3 and a rea...

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