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3 years ago

Consider the equilibrium system below.

Chemistry

2 answers:

Katarina [22]3 years ago

5 0

Answer:

It's C, the rate of sublimation of the NH4Cl crystals is equal to the rate of solidification of the NH4Cl vapors.

Explanation:

Neporo4naja [7]3 years ago

3 0

Answer:

C. The rate of sublimation of the NH₄CI crystals is equal to the rate of solidification of the NH₄CI vapors.

Explanation:

The question is garbled and the reactants and products do not match.

This is the correct question, assuming the products are correctly described:

"Consider the equilibrium system below.

$NH_4Cl(s)\rightleftharpoons NH_4Cl(g)$

If the system is at dynamic equilibrium, which statement is true?

A. Sublimation of the NH₄CI crystals stops.

B. Solidification of the NH₄CI vapors stops.

C. The rate of sublimation of the NH₄CI crystals is equal to the rate of solidification of the NH₄CI vapors.

D. The rate of sublimation of the NH₄CI crystals is higher than the rate of solidification of the NH₄CI vapors"

<h2>Solution</h2>

The left side of the equation shows solid NH₄CI. The right side shows gas NH₄CI.

Thus, the equation represents the equilibrium between the formation of crystals (solid) and vapors (gas) of the compound NH₄CI.

The change from solid state to gas state is represented by the forward reaction (fom left to the right). It is named sublimation.

The change from vapors to crystals (named deposition) is represented by the reverse reaction (from right to left).

Dynamic equilibrium means that the molecules continually form vapors and cristals at the same speed. Thus, the final result is that each the total number of molecules of vapors and the total number of molecules of crystals do not change.

In conclusion, "The rate of sublimation of the NH₄CI crystals is equal to the rate of solidification of the NH₄CI vapors" (option C).

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frutty [35]

Answer:

The value of $K_p$ at 4224 K is 314.23.

Explanation:

$O_2(g)\rightleftharpoons 2O(g)$

Initially

4.97 atm 0

At equilibrium

4.97 - p 2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = $p_{O_2}=0.28 atm$

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$K_p=\frac{(p_{O})^2}{(p_{O_2})}$

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3 years ago

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Andru [333]

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3 years ago

For parts a & b below, derive only the initial value problem set up.

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Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

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Now; y(0) = 25

25 = Ce⁰

C = 25

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