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ratelena [41]
3 years ago
5

Consider the reaction 2 SO2(g) + O2(g) <=> 2 SO3(g), which is exothermic as written. What would be the effect on the equil

ibrium position of removing O2(g)?
Chemistry
1 answer:
enot [183]3 years ago
5 0

Answer:

Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

Explanation:

The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.

The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.

Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
Mazyrski [523]

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>

I hope it helps!

6 0
3 years ago
Description of golgi bodies
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8 0
3 years ago
Read 2 more answers
1. In an equilibrium experiment, acetic acid (which is a weak acid) is mixed with sodium acetate (a soluble salt), with methyl o
koban [17]

Explanation:

It is known that acetic acid is a weak acid. It's equilibrium of dissociation will be represented as follows.

          CH_{3}COOH(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)

On the other hand, sodium acetate (CH_{3}COONa) is a salt of weak acid, that is, CH_{3}COOH and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.

Since, acetic acid is a weak acid but still it is an acid. So, when methyl orange is added in a solution of acetic acid then it given a reddish-orange color because of its acidity.

When sodium acetate is mixed into this solution then it will dissociate as follows.

            CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)

As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.

So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.

As a result, there will be decrease in the concentration of hydronium ions. As a result, there will be increase in the pH of the system.

Hence, color of methyl orange will change from reddish orange to yellow. This shift in equilibrium is due to the common ion which is CH_{3}COO^{-} ion.

4 0
3 years ago
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3 years ago
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What are some physical properties of a candle
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The wick and the wax

Sorry if that was useless, I'm not sure how generalized you were being

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