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devlian [24]
2 years ago
6

For each event stated below, indicate how the concentration of each species in the chemical equation then change to reach equili

brium. Click in the blue box below to toggle the relative change in concentration (up=increases, down=decreases leave the box blank for no change). 2CO(g)+O2(g) 2CO2 (g) increasing the concentration of CO increasing the concentration of CO2 decreasing the the volume of the system
Chemistry
1 answer:
xxTIMURxx [149]2 years ago
7 0

Answer:

Answers are in the explanation

Explanation:

It is possible to know the effect of change the conditions in an equilibirum based on LeChatelier's principle that says any change in conditions of a chemical system produce in the system an opposite response to counteract the initial change.

For the reaction:

2CO(g) + O₂(g) ⇄ 2CO₂(g)

<em>Increasing the concentration of CO: </em>The system will response producing more CO₂ counteracting the increasing of CO. CO concentration decreases, O₂ concentration decreases, CO₂ concentration increases

<em>Increasing the concentration of CO₂: </em>Here, the system response producing more CO and O₂ decreasing the concentration of CO₂

<em>Decreasing the volume of the system</em>: Here, the system will try to go in the direction that occupy less volume to counteract the increase of pressure because of decreasing of volume. In the left of the reaction there are three moles of gas and in the right there are just two. That means the system will produce more CO₂ decreasing, thus, concentration of CO and O₂.

I hope it helps!

<em />

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Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

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