[romick]

How is going to change the speed of reactions, if the volume of every system will decrease  4 times:

a) S+O2=SO2
b)2SO2+O2=2SO3

[sjb]

That will depend on the nature of the rate determining step, which is not always apparent from the balanced equation.

[romick]

How is going to change the speed of reactions, if the volume of every system will decrease  4 times:

a) S(s)+O2=SO2(g)
b)2SO2(g)+O2(g)=2SO3(g)

We know 1) Cm=ν/V ,and speed is 2) v=kCmACmB, so if the Volume decrees 4 times than from 1) Cm increes 4 times.
then,
a) v1=kCmB , v2=k(4CmB) , v2/v1=k4CmB/kCmB=4 The speed of reaction a) S(s)+O2=SO2(g) will increase 4 times if the Volume  will decrease 4 times.
b)v1=k[(CmA)^2]CmB , v2=k[(4CmA)^2]4CmB , v2/v1=k[(CmA)^2]CmB/k[(4CmA)^2]4CmB=64 The speed of reaction b)2SO2(g)+O2(g)=2SO3(g) will increase 64 times if the Volume will decrease 4 times.
                                                                                                                                      R: a)4;b)64.
This is result i was expecting for, thanks for nothing, already solved it by myself. 

[sjb]

Sorry, can you explain what things like Cm, ν, V are in these equations - I may have misunderstood what you mean in the initial question

[renge ishyo]

He's trying to deduce the behavior of systems from simple mathematical considerations. For example:

A + B ---> C

If this was the rate determining step, and the only step in the reaction, then you can write the rate law like this:

rate = k{A}¹{B}¹

Now if you decrease the volume of the system, the values of both {A} and {B} would increase because they are defined as mass/volume. So the smaller the volume with the same mass, the bigger the concentration, and so using the law alone you would predict the rate would increase (and if it happens to increase the rate, the effect could be described as arising by increasing the number of collisions in a given amount of time by decreasing the volume).

The problem with the analysis is the conclusion neglects the fact that the rate constant "k" can change with the conditions. Decreasing the volume will normally change either the pressure or the temperature depending on the circumstances, and this can affect the rate constant substantially. So the answer could really be anything, it can speed up, slow down, show no change, or even stop depending on the effect the change in the system has on the rate constant k. It is for this reason that chemists always say that you cannot derive the kinetics of a reaction analytically. The results must come from experiment.

[sjb]

Ahh, so we were talking about the same thing really.

[romick]

Sorry; sure, i should've told that this is for school level chemistry not for real, i solved it as they taught me in school, but you really good explained it how it will go, if take an experiment; thanks for that.