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What determines a reversable and irreverasable reaction? Is it related to non-communitive properties or what?


Theoretically, all reactions are reversible, but for all practical purposes this is not the case.

A reversible reaction is one where the equilibrium is so close that... well... there really is no equilibrium.

I'm having a rather hard time explaining this. :) , so if someone else wants to step in...

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Ok well equillibrium chemistry is hardly 101, but the basics are that reactions happen in both directions - forwards and backwards.


For example, the haber process (ammonia production) is as follows:

[math]N_2(g) + 3H_2(g) <-> 2NH_3 \Delta H = -92kJ/mol[/math]

which can be re-written as:

[math]2NH_3 <-> N_2(g) + 3H_2(g) \Delta H = +92kJ/mol[/math]


You may be tempted to think that if you combine 3 parts H with one part N you will produce 2 parts ammonia, but that is not the case, at the same time that you are creating NH3 out of H and N, NH3 is also decomposing to form H and N. There is a difference in the rate at which these two reactions happens, but eventually a dynamic equillibrium is reached and the concentrations of the three chemicals remains stable. Which reaction is favoured is called the equillibrium position. Note this does not mean the reaction has ceased, just that for the given concentrations and reaction rates the overall concentrations remain the same.


Some other equillibrium concepts for those interested:

You may have noticed that the reaction is delta H -ve one way and +ve the other. This means energy is released in the -ve reaction and energy is stored in the +ve one. The equillibrium will be affected by the temperature of the surroundings, for example, if you where to increase the temperature then the reaction that absorbs the heat (+ve [math]\Delta[/math]H reaction) will be favoured and hence there will be a shift in the concentrations of the equillibrium mixture. In this case a raise in temp will cause the mixture to have higher concentrations of N and H. While if you lower the temp the opposite will happen and through much the same reasoning NH3 will be present in higher concentrations.

Pressure will also affect the equillibrium position, you may have noted that one side of the equation has 4 moles worth of gas and through the reaction only produces 2 moles of gas. Therefore, if you where to increase the pressure of the system then the reaction has more molecules around is likely to precede faster, and hence shift the equillibrium so that the pressure will be reduced.


Another important thing to note is that catalysts do not change an equillibrium position. Catalysts provide an alternative pathway for the reaction to happen that requires less activation energy, hece speeding it up. This will work both ways, the forward and backward reaction will both be sped up and hence the equillibrium position is maintained.


There is handy way of thinking out these concepts, its called Le Chateliers principle. It states the equillibrium will be shifted to counteract the change made to the system. It works fine by reason, but be carefull, some are tempted to say (and erroneously at that) the system counteracts a change, the system doesnt give a rip, the conditions of the surroundings just put it is such a way as to favour the opposite reaction. I always favour the method of working through it via logic alone.


For this reason I was adament to accept that there is such a thing as an irreversable reaction.. not in chemistry at any rate, for I would not consider a nuclear reation or anti-matter reaction to be part of conventional chemistry.

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Acid/base equilibrium hey!


Well I guess I should introduce the idea of an equilibrium constant, Kc. It is the ratio of the product of the concentrations of the products over the product of the concentrations of the reactants! which is probably better summarised in an equation rather than words, which sticking with the haber process like before:


[math]N_2(g) + 3H_2(g) <-> 2NH_3(g)[/math]


[math]Kc = \frac{[NH_3]^2}{[N_2][H_2]^3}[/math] the units of which are worked out by cancelling the concentration units. In this case it would be [math]M^{-2}[/math]. The equilibrium constant can only be calculated once the equilibrium has been reached, otherwise it would be in error.


Now you are already familiar with what may be considered a strong or weak acid, this is what makes them strong or weak - the extent to which the reaction occurs.


When an acid is in solution the acid donates a proton very easily to water to make the hydronium ion.


eg water will sefl ionise itself [math]2H_2O(l) <-> H_3O^+(aq) + OH^-(aq)[/math]

which gives rise to:

[math]Kc = \frac{[H_3O^+][OH^-]}{[H_2O]^2}[/math]

water can be excluded from this equation because its concentration is always 1, hece:

[math]Kc = [H_3O^+][OH^-][/math]

The Kc for this reaction at room temp is [math]10^{-12}[/math]


The pH of a solution is said to be:


[math]pH = -log([H_3O^+])[/math]


There is some very interesting equilibrium reactions going on in your body! One is the system between Heamoglobin, oxygen and carbondioxide in your respiritory system.


I might get round to posting some more later, but right now that will have to do :)



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They are the same chemically insofar as that they are both surfactants designed to reduce the surface tension of water so things will get wet better. But a detergent will also have other chemicals that help clean that could well be harmfull to soil.


Mhmm. Chemical wetting agents are designed specifically for that purpose.

Detergents are generally... well... cleaing agents.

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  • 1 month later...

Polar molecules have elemental components that are arranged in a manner that provides a "negative end", and a "positive end".

Water is one such molecule. The hydrogens have a slight positive charge, and the oxygen has a slight negative charge.

This is especially important in biology for multiple reasons. For instance, polar molecules are eoften very good solvents.

You can ask our user HydrogenBond for more info. :cup:

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