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1 year ago

A container of N, 03(9) has a pressure of 0.490 atm. When the absolute temperature of the N, O2(g) is tripled, the gas

Chemistry

1 answer:

spayn [35]1 year ago

3 0

Answer: 1.59atm

Explanation:

We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."

it can be said that

The final pressure of the gas mixture, assuming that the container volume does not change =

From the question we are told

A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g).

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Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

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(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

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As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

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