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#Post#: 1269--------------------------------------------------
Oxidation Number, and understanding Le Chatelier principle
DIR By: Felecity2394
Date: November 27, 2012, 2:30 pm
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Can someone please help me understand the difference between
reducing agent and oxide agent, i'm a little confused on it. I
tried using O.I.L.R.I.G. (Oxidation is Losing, Reduction is
Gaining). But I don't know why its not helping me.
Another question that I have is pertaining to Le Chatelier
Principle.
On my review worksheet it states: Understanding Le Chatelier's
principle: the effect of the following factors on the position
of Equilibrium:
a) Change in concentration of reactants or products
b) Change in temperature
c) Change in pressure
for each of the letter can someone give a equation and explain
each concept. Please and Thank You :)
#Post#: 1271--------------------------------------------------
Re: Oxidation Number, and understanding Le Chatelier principle
DIR By: Shiva
Date: November 27, 2012, 11:56 pm
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Hiya, welcome to Chemistry Planet :-)
Technically, the definition is yes, Oxidation is losing
electrons. To make it happen, we need an oxidizing agent and it
has to gain electrons for successful oxidation. The reduction is
opposite.
But, you need not to look at it technically everytime to
understand it. Oxidizing agents are something that can be
reduced easily. They are always hungrier for electrons.
For example,
S + 6HNO3 = H2SO4 + 6NO2 + 2H20
Nitric acid is a strong oxidizing agent because it can be
reduced easily. In this reaction, nitric acid oxidizes sulfur to
sulfuric acid and itself reduced to nitrogen dioxide and water.
Also, some people assume whatever contains oxygen are oxidizing
agents and more oxygen means it is a strong oxidzing agent. But,
it is completely a wrong statement. There are oxidizing agents
without having oxygen atoms. For example, Chlorine. Sulfates
have 4 oxygen atoms but they are not oxidizing agents because
they can't be reduced easily and not ready to gain any electrons
either. However, some sulfates are reasonably good reducing
agents. For example, Ferrous sulfate. But, some oxidizing agents
react differently in alkaline conditions and acidic conditions.
For example, Hydrogen peroxide. When two strong oxidizing agents
react, always the stronger ones win. Below is the good example
for it:
O3 + H2O2 = 2O2 + H20
Here, Ozone is an aggressive oxidizing agent and hydrogen
peroxide can't even stand against it. It has to lose here to
Ozone.
Hope this is clear. Please let me know if you have any
questions.
Plz can anyone answer Le Chatelier's part?
Shiva
#Post#: 1272--------------------------------------------------
Re: Oxidation Number, and understanding Le Chatelier principle
DIR By: Michel
Date: November 28, 2012, 5:01 am
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Probably it'll be nice if we take them one after the
other...Felecity, are you clear on the Redox part? If you have
more questions please ask..
#Post#: 1274--------------------------------------------------
Re: Oxidation Number, and understanding Le Chatelier principle
DIR By: Cheminized
Date: November 28, 2012, 8:10 am
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Felecity, welcome to Chemistry Planet :-)
i think what Shiva said about redox is pretty strait forward ,
and if you still have questions about it the staff will be there
for you.
now considering Le Chatelier's principle , there is one thing
that you need to know about it :
the equilibrium will move in reverse with respect to the factor
that has changed, i know its confusing but i will explain with
examples :
a) Change in concentration of reactants or products:
A+B===C
if we were to increase the concentration of the reactants (ie A
and B)
the equilibrium will move to tolerate the increment of the
reactants, and that can be achieved by making more of the
products, then the equilibrium will move to the right.
like wise if we increase the amount of products the system will
make more reactants by dissociating the product C to form the
products a gain so the equilibrium will move to the left.(hope
that's clear).
b) Change in temperature:
we have to kinds of reactions (thermally speaking):
endothermic reactions (reactions that NEED thermal energy to
stay a life)
exothermic reactions (reactions that GIVE thermal energy to
stay a life)
for endothermic reactions lets consider heat as an imaginary
reactant (since the reaction need it)
and for exothermic reactions lets consider heat as an imaginary
product , having that said we can apply what we said about the
change in the products and reactants:
c) Change in pressure
that is important for reactions that involves gaseous species:
if we increase the pressure the equilibrium will move to the
side that will decrease the pressure, and decreasing the
pressure of the system will move the equilibrium to the side
that will increase the pressure:
A(g) +B(g)===== C(g)
for this reaction we have more gaseous molecules on the left
side so going to right will decrease the pressure of the system
and visversa for for going to the left so if increase the
pressure of the system the equilibrium will go left and
viseversa for decreasing the pressure of the system.
in a nut shell moving the side that have more moles of gaseous
species will result in an increment of the system pressure.
in a smaller nut shell increasing the pressure of a system will
move the equilibrium to the side that have less moles of gaseous
species. and decreasing the pressure of the system will move the
equilibrium to the side that have more moles of gaseous species
.
hop that was miss leading.
feel free to ask >>>>
#Post#: 1276--------------------------------------------------
Re: Oxidation Number, and understanding Le Chatelier principle
DIR By: Shiva
Date: November 28, 2012, 8:33 am
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Thank you Chem :-)
#Post#: 1346--------------------------------------------------
Re: Oxidation Number, and understanding Le Chatelier principle
DIR By: Cheminized
Date: December 1, 2012, 12:58 pm
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Welcome bro. ;D
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