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liberstina [14]
3 years ago
7

When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.

1 atm . Some of the CO2 decomposes to CO and O2.
Calculate the mole percent of CO2 that decomposes.
Express your answer using two significant figures.
Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

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Answer:

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Explanation:

Let's say the reaction is

R ⇌ 2P; endothermic

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Another way to write the equilibrium would be

heat + R ⇌ 2P

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Let's consider each of the stresses in turn.

(i) Changing the temperature

If you want to increase the amount of product, you increase the temperature. The system will try to get rid of the added heat by shifting to the right, thus forming more product.

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If R and P are liquids or solids or in aqueous solution, changing the pressure will have no effect. Something must be in the gas phase for a change in pressure to affect the position of equilibrium.

If P is a gas, the equilibrium is

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Under what conditions of temperature and pressure is a gas least soluble in water
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The conditions of temperature and pressure in which a gas least soluble in water is low pressure and high temperature.

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The amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid, according to Henry's law.

From this law it is clear that:

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5 0
3 years ago
An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00
valkas [14]

Answer:

  • <u>Option A. 450.00</u>

Explanation:

<u>1) Data:</u>

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

<u>2) Principles and equations</u>

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

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Where,

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<u>3) Solution:</u>

<u>a) Substitute the data into the equation:</u>

  • 267.3 kJ = 18 kg × S × (318 K - 285 K)

<u>b) Solve for S and compute:</u>

  • S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

<u>c) Convert to J /( kg . k):</u>

  • 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

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