[TKA]

First of all i´m new to the forum so hello everyone. I´m having problems with a chemistry exercise and i think this is the right forum. I would like some help with this:

"Because of the stoichiometry , the progress of the reaction maybe different for  the different substances that are participating in the process. We use the variable z to uniform this progress. For example, in the hypotethic reaction 2R P, were consumed 2z mol of R, so in the same time were obtained z mol of P. Complete the table, knowing that n_a= 20-3z, with the values of  :delta: n per time and the values of  :delta: z. Estabilish a mathematical relashionship between  :delta: n_a and  :delta: z,  :delta: n_b and  :delta: z"

Time(min) 3A          2B              z           :delta: n_a    :delta: n_b   :delta: _z 

0         20 mol         0 mol              0            #       #       #
10       11 mol         6 mol              3           -9       6        3
20        5 mol         10 mol             5           -6       4        2
30        2 mol         12 mol             6           -3       2        1


I completed the table (bold) acoording to the following:

n_b= 0-2z    n_b= -2z

 :delta: n_b=n_bf-b_bi = (+ or -?)2z-0= (+ or -?)2z
 :delta: n_a=n_af-a_ai = 20-3z-20 = -3z
 z = -(:delta: n_a/3)  or z =  :delta: n_b/2


Is my reasoning correct? In this case, z =  :delta: z  so i can switch in the equation?  It is -2z or 2z?

Ps: I have more doubts, but an exercise "pulls" the next one, so i need to go step by step to proceed.

If someone can help me, i would aprecciate that.
Sorry for my english (anything), i´m not a native speaker.

[Hunt]

Hello TKA,

z = -(:delta: n_a/3)  or z =  :delta: n_b/2


Is my reasoning correct? In this case, z =  :delta: z  so i can switch in the equation?  It is -2z or 2z?

Ps: I have more doubts, but an exercise "pulls" the next one, so i need to go step by step to proceed.

If someone can help me, i would aprecciate that.
Sorry for my english (anything), i´m not a native speaker.

Your result is correct.

In general , for any chemical reaction with r chemical species , there exists r variables. From the stoichiometry you can write :

dn₁ / v₁ = dn₂ / v₂ = . = dn_r / v_r

where v_i is the stoichiometry number. It is negative for reactants and positive for products. The reason is that the reactant particles are decreasing while the product particles increase. ( For e.g. in your problem you write :  :delta: n_a = - 3z < 0 , :delta: n_b = + 2z > 0 )


So from the r variables, there are r-1 dependent variables and only one independent variable, usually denoted zetta the extent of the chemical reacntion. ( called z in this problem )

You can then simply write :

 :delta: n₁ / v₁ =  :delta: n₂ / v₂ = . =  :delta: n_r / v_r = z

Any chemical reaction can then be quantitatively followed by a single independent variable, which is taken here to be positive. Should you choose to take z < 0 , then your equations are inverted here , this is not a standard thing , however.

For a more general discussion , see for e.g. "Modern Thermodynamics, From Heat Engines to Dissipative Structures" for Ilya Prigogine , chapter 2 , section 2.5 "Extent of a Reaction : A State Variable for Chemical Systems"

[TKA]

Ok, i think i understand what you said, thank you very much.

This is where we started to get stuck.

"The dimerization reaction for NO₂ is: 2NO₂  N₂0₄. :delta: G_Fº(NO₂)= 51Kj/mol,  :delta: G_Fº(N₂O₄)= 98Kj/mol. Considering that G_r (free energy of the system) depends only in the quantity of substance (n) and the values for :delta: G_Fº, i.e. G_r=n_NO2.:delta: G_Fº(NO₂) + n_N2N04.:delta: G_Fº(N₂O₄)  and suposing the proccess beginning with 20 mols of reactant and 0 mols of product and in the end there is just product (100%) write G_r=f(z) and layout the graph G_r versus z. Using the data for the end state and the start state, calculate  :delta: G_r in the conditions previously described and also calculate  :delta: G/ :delta: z (what relationship this represents in the graph?). Find a expression to :delta: G_r=f( :delta: z)."


As you can see, now its a little complicated.

For 
Gr_f = 0 + 10 .98 = 980 Kj/mol
Gr_i = 20. (51) + 0 = 1020 Kj/mol
 :delta: G_r = 980 - 1020 = -40Kj/mol

Gr = f(z)
Gr = (20 - 2z).:delta: G_Fº(NO₂) + (0-z).:delta: G_Fº(N₂O₄) (as you said i will have a positive z here, so...)
Gr = 1020 - 102z + 98z
Gr = 1020 - 4z (is this correct?)

If this is correct the graph Gxz will be a straight line?
So here, i will have  :delta: z = 10 and  :delta: G /  :delta: z = -40/10 = -4?

What would mean the relationship between  :delta: g/ :delta: z (including the graph).
I also having dificulties to find the last expression asked.


I am missing something or i don´t know..

ps: I apreciate your book suggestion and already research for it, but i live in Brazil and don´t have access...

[Hunt]

You have defined delta Gr(z) in a wrong way.

Gr = (20 - 2z). GFº(NO2) + z GFº(N2O4)

You added a minus sign before GFº(N2O4), which is wrong. I'll write the correct expressions in latex and post them in a .pdf.

[TKA]

Ok, this LATEX is a "program" or just the plugin for the forum?


Regards,

[Hunt]

LATEX is a language used to create mathematical documents. For example this .pdf is generated in latex. This forums supports latex but I forgot its command lines...

There are a couple of things about the problem which do not make sense , but I won't mention them to avoid any confusion for you. I will keep track of the topic if you wish to discuss anything more about this.