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Gre4nikov [31]
3 years ago
5

Given the reaction system in a closed container at equilibrium and at a temperature of 298 K:

Chemistry
2 answers:
Tomtit [17]3 years ago
7 0
Constant. The concentrations must be constant and the rates must be equal.
MAVERICK [17]3 years ago
4 0

Answer:

The measurable quantities of the gases at equilibrium must be constant

Explanation:

Measurable quantities for the gases at equilibrium includes rate of formation of NO_{2}, rate of formation of N_{2}O_{4}, pressure of NO_{2} and pressure of N_{2}O_{4}.

At equilibrium, rate of forward reaction is equal to rate of reverse reaction.

Rate of forward reaction infers rate of formation of NO_{2} and rate of backward reaction infers rate of formation of N_{2}O_{4}.

As rate of formations from both direction are equal therefore pressure of both gaseous species remains constant.

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This system has an equilibrium constant of 0.105 at 472°C: N2(g) + 3H2(g) ↔ 2NH3(g)
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Read 2 more answers
Consider the following system at equilibrium:
alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0
3 years ago
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