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nirvana33 [79]
3 years ago
11

Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine

Chemistry
1 answer:
andreev551 [17]3 years ago
3 0
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
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Acetic acid and ethanol react to form ethyl acetate and water, like this:
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Answer:

1.) Option C is correct.

The rate of reverse reaction is greater than zero, but equal to the rate of the forward reaction.

2) Option B is correct.

The rate of reverse reaction is Greater than zero, but less than the rate of the forward reaction.

3) Option C is correct.

The rate of reverse reaction is Greater than zero, and equal to the rate of the forward reaction.

4) Option A is correct.

How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium? Zero.

Explanation:

HCH,CO2(aq) + C2H5OH(aq) ⇌ C2H,CO2CH3(aq) + H2O

1) Before the main product is removed from the reaction setup, the chemical reaction is at equilibrium.

Chemical equilibrium is a state of dynamic equilibrium such that the concentration of the reactants and the products do not always remain the same but the rate of forward reaction always matches the rate of backward reaction.

2) When 246. mmol of C2HCO2CH3 are removed from the reaction mixture....

And when one of the factors involved in chemical equilibrium changes, Le Chatellier's principle explains that the system then adjusts to remedy this change and takes time to go back to equilibrium again.

When one of the species involved in the chemical reaction at equilibrium, is removed from the reaction mixture, the rate of reaction begins to favour that side of the reaction until equilibrium is re-established.

So, when 246 mmol of one of the products is removed, the response is to cause the rate of forward reaction to be favoured to produce more of products as there are fewer, and the rate of reverse reaction at this moment becomes less than the rate of forward reaction.

3) The rate of the reverse reaction when the system has again reached equilibrium

Like I said in (2) above, the reaction remedies this change in concentration of one of the products until equilibrium is re-established and when chemical equilibrium is re-established the rate of forward reaction once again matches the rate of backward reaction.

4) How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

By the time equilibrium is re-established, the system goes back to how it all was and the concentration of C2H5CO2CH3 goes back to the same as it was at the start of the reaction.

Hope this Helps!!!

3 0
3 years ago
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