luv's G-Chem Question

[luv8724]

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A Force is applied to a container of gas reducing its volume by half. The temperature of the gas:
a) decreases
b) increases
c) remains constant
d) The termperature change depends upon the amount of forced used

Second Question

4NH3(g) + 5O2(g) - 4NO(g) + 6H2O(g)

Which of the following would be true if the container were allowed to expand at constant temperature?
a) Initially during the expansion the forward reaction rate would be greater than the reverse reaction rate
b) The equilibrium would shift to the left
c) The partial pressure of oxygen would increase
d) The pressure inside the container would increase

 

[jh311]

...

 

[UCB05]

The use of force to reduce volume is a net input of energy. This will manifest partly as an increase in temperature of the gas. The force required to reduce the gas to half volume will be the same done slowly or quickly, only varying in power, so d is false. This is the concept behind diesel engine ignition and fire pistons. The sudden forceful decrease of volume in a cylinder increases the temperature inside enough to ignite a combustible substance inside the chamber.

This question is different from one that asks if a gas shrinks to half volume by itself at constant pressure, in which case charle's law states that temperature decreases proportionately to volume decrease.


2) The ratio is 9 mols of gas on the reactant side for every 10 mols of gas on the product side. An increase in pressure or decrease in volume will favor the side with fewer mols of gas. An increase in volume at constant temperature would then favor the side with more mols of gas. This appears as a transitory increase in forward reaction rate (A) during the expansion, before equilibrium is reestablished. By Boyle's law, pressure would have to decrease with increased volume.

Even if you only had a rudimentary understanding of LeChat's principle, you could deduce the answer. A shows a shift toward the products. B states a shift toward reactants. C represents a shift toward reactants (increase in O2 mol fraction, which must be accompanied by decrease in products' mol fractions). Pick the odd man out and go with A.

 

[IdahoDoc]

The use of force to reduce volume is a net input of energy. This will manifest partly as an increase in temperature of the gas. The force required to reduce the gas to half volume will be the same done slowly or quickly, only varying in power, so d is false. This is the concept behind diesel engine ignition and fire pistons. The sudden forceful decrease of volume in a cylinder increases the temperature inside enough to ignite a combustible substance inside the chamber.

This question is different from one that asks if a gas shrinks to half volume by itself at constant pressure, in which case charle's law states that temperature decreases proportionately to volume decrease.


2) The ratio is 9 mols of gas on the reactant side for every 10 mols of gas on the product side. An increase in pressure or decrease in volume will favor the side with fewer mols of gas. An increase in volume at constant temperature would then favor the side with more mols of gas. This appears as a transitory increase in forward reaction rate (A) during the expansion, before equilibrium is reestablished. By Boyle's law, pressure would have to decrease with increased volume.

Even if you only had a rudimentary understanding of LeChat's principle, you could deduce the answer. A shows a shift toward the products. B states a shift toward reactants. C represents a shift toward reactants (increase in O2 mol fraction, which must be accompanied by decrease in products' mol fractions). Pick the odd man out and go with A.

👍 You got it not sure what the other person above your post was talking about

 

[luv8724]

UCB You got it correct 😛

but can you explain a little bit more why increase in O2's partial pressure would shift the reaction to the left?

 

[CANgnome]

increase in o2's partial pressure can only be possible if the volume decreases? so if o2 increased partial pressure, it would mean that the equation shifted to the left, with a shrinking volume?

 

[UCB05]

UCB You got it correct 😛

but can you explain a little bit more why increase in O2's partial pressure would shift the reaction to the left?

Increase in pO2 doesn't shift the reaction left. The increase in O2 would be a result of a shift to the left. partial pressures are directly proportional to mol fractions. If the partial pressure of O2 increases, more O2 has to have been produced, and looking at the equation, all the oxygen must have come directly from the products.

(more reactants & less products) = shift to the left.

*edit*

increase in o2's partial pressure can only be possible if the volume decreases? so if o2 increased partial pressure, it would mean that the equation shifted to the left, with a shrinking volume?

Yes. Don't confuse cause and effect. In this question, the shrinking container causes the equilibrium to shift. That option just requires thinking a step further. Decrease volume = more reactants = more O2 = increase in pO2. Since the question states in increase in volume, that option is false.

 

[luv8724]

Increase in pO2 doesn't shift the reaction left. The increase in O2 would be a result of a shift to the left. partial pressures are directly proportional to mol fractions. If the partial pressure of O2 increases, more O2 has to have been produced, and looking at the equation, all the oxygen must have come directly from the products.

(more reactants & less products) = shift to the left.


oh okay I see what you are saying UCB. I kept thinking that by increasing the temperature would no matter what favors towards the reactants since a lot of the review book has this example as in exothermic. Of course it will favor the products in trying to increase entropy of the reaction. I dont know if I'm getting it right but I get the concepts u r saying.

 

[UCB05]

Well the second question gives you the reaction, but says constant temperature, so it's not a factor...