Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: 777888 on November 18, 2004, 10:56:39 PM
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Hi, I have some questions!
* = is an double arrow
1) CO2 + 4H2 = CH4 + 2H2O + heat
Predict, using Le Chatelier's principle, the conditions required in a closed system to produce MAXIMUM amount of methane.
2) CaCO3 + 2H+ = Ca2+ + H2O + CO2
How would the hardness of the water [Ca2+] affect the growth of stalagmites and stalacites (ie CaCO3)?
[For this one, I don't even get what the question is asking...]
3)CuCl4(2-) + 4H2O = Cu(H2O)4(2+) + 4Cl-
Predict the shift and effects in the equilibrium after AgNO3 is added.
Thank you! :)
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What are your thoughts?
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What are your thoughts?
#1)-decrease temperature, increase concentration of CO2 and H2, decrease concentration of H2O, decrease volume... am I right?
#2) no.2 I don't understand the question
#3) there are no Ag+ and NO3- in the reaction, so I don't really know how to predict...
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1) CO2 + 4H2 = CH4 + 2H2O + heat
Predict, using Le Chatelier's principle, the conditions required in a closed system to produce MAXIMUM amount of methane.
u have the correct idea, but you should say "low temperature" and not "decrease temperature". You are supposed to describe the condition, not what you are suppose to do.
5moles of gas reactants produce 3 moles of gas reactants (assuming the reaction condition is such that water is in vapour phase). This implies high pressure would favour the forward reaction. Incidentally, this would suggest an increase in concentration [partial pressure] of the reactants and a smaller volume for the reaction system to take place in. Since the forward reaction is exothermic, a low temperature environment would favour the forward reaction. All this are in accordance to Le Chatelier's Principle.
2) CaCO3 + 2H+ = Ca2+ + H2O + CO2
How would the hardness of the water [Ca2+] affect the growth of stalagmites and stalacites (ie CaCO3)?
the higher the hardness of water, the more aq. Ca2+ present. This should provide sufficient hint to your question here.
3)CuCl4(2-) + 4H2O = Cu(H2O)4(2+) + 4Cl-
Predict the shift and effects in the equilibrium after AgNO3 is added.
What happens when you add aq silver nitrate to a chloride solution?
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Thanks a lot!
So #2, the equilibrium will shift to the left if there is a higher hardness of water.
#3 I don't really know! Single displacement? Will Cl- ions reaction with AgNO3?
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The Cl- ion react with Ag+ from AgNO3 to form a white ppt.
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For question number 1, being able to remove the water from the system would also be a good maneuver. Now if the reaction conditions are at a low temperature, the water should be condensing into liquid again after it is generated. To prevent water vapor from coming off of the liquid water, you could put an anhydrous cobalt compound in there to absorb the water as it's created and lock it up in a crystalline structure. This will also increase the amount of methane produced.
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LOL. In a chemical plant, the equilibrium mixture will be sent into a seperator whereby most water will be removed as effluent, methane as a pure stream, and the rest back into the reactor to recycle the reactants for better economy.
Upon addition of aq AgNO3, the chloride in solution will be precipitated out as AgCl (s), thus reducing the amount of Cl- present. This favours the forward reaction of the equillibrium below:
CuCl4 2- (aq) <-> Cu2+(aq) + 4Cl-(aq)
Some green CuCl4 2- (aq) complex will be converted to Cu 2+ (aq). Hence, the solution turns blue and white precipitate (AgCl) is formed.
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"Upon addition of aq AgNO3, the chloride in solution will be precipitated out as AgCl (s), thus reducing the amount of Cl- present"
Sorry... I don't get why would this happen...
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this is because AgCl has a very low solubility product constant. a low concentration of chloride ions is sufficient to cause precipitation of AgCl in the presence of aq Ag+ ions
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For the equilbirium reaction:
PCl5 = PCl3 + Cl2
Say initially there is 1.00 mol of PCl5, why would the equilibrium concentration of PCl5=0.80mol/L, of PCl3=0.20mol/L, of Cl2=0.20mol/L? In a graph, how do you know when does PCl5 stop decreasing and reach equilibrium? Can this be calculated? Besides, why would the equilibrium concentraion of PCl5 equal to that of Cl2(ie 0.20mol/L)?
If I start with PCl3 and Cl2 OR start with 3.00 mol of PCl5, would the equilibrium concentration of each entities be the same as the one above? (ie PCl5=0.80mol/L, of PCl3=0.20mol/L, of Cl2=0.20mol/L)
Thank you!
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Given the equilibirum constant and the equilibrium expression, you should be able to manipulate the algebreic expression to get the required values. The table below should help:
PCl5 <=> PCl3 + Cl2
for a Fixed Volume 1-Litre Equilibrium Mixture,
| | PCl5 | | PCl3 | | Cl2 |
initial amount: | | 1.00 | | 0 | | 0 | |
change in amount: | | -x | | +x | | +x |
equilibrium amount: | | (1.00 - x) | | x | | x |
Note, the amount reacted agrees with the stoichiometric ratio, ie. X moles of PCl5 breaks down to form X moles of PCl3 and X moles of Cl2
the equilibrium expression is:
K = [PCl3][Cl2]/[PCl5] = x2 / (1.00-x) where K is the equilibirium constant
This expression can be manupilated to the form of a quadratic equation which you should be able to solve
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Help, I have a really hard one!
H2 + I2 <-> 2HI
Predict the effects(increase or decrease in concentration) if N2 is added!
What would happen? ???
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ask yourself 2 questions:
1. does nitrogen react with any equilibrium component?
2. does addition of nitrogen increase the volume of the system?
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ask yourself 2 questions:
1. does nitrogen react with any equilibrium component?
2. does addition of nitrogen increase the volume of the system?
1. Will N2 react with H2? How can I know?
2.No
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if there is no change in volume, the concentration of hydrogen gas remains the same.
nitrogen doesn't react with hydrogen, unless lotsa of heat is provided, It takes up a lot of activation energy just to break the N-N triple bond.
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So is N2 acting as an INERT gas to the system?
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nitrogen is an inert gas in the system.