Chemical Kinetics: velocity and equilibrium

[dRic2]

Hi everybody,

Given a generic reaction $Reagents ↔ Products$ the velocity of the reaction is $R = R_{right} - R_{left}$

To take in account that the velocity of the reaction will slow if approaching the equilibrium we say that
$$ R = R_{right}(1 - \frac K K_{eq}) $$
Although it seems reasonable it appears to me too much "arbitrary" so I don't get it very much. I mean, I could have elaborated an other relation to take the equilibrium into account or is it the only one?

Second question: A book of mine says that I reaction of first order such as $A → B$ has
$$ R = kC_A(1- \frac K K_{eq} C_B) $$
but it seems wrong to me... Wouldn't it be
$$R = R_{right} - R_{left} = R_{right}(1 - \frac K K_{eq}) = kC_A(1- \frac K K_{eq} ) $$ ?

 

[mjc123]

Your book equation looks wrong dimensionally. Check their definition of the various k's.
Can you define the terms you use? Given those definitions, can you use the 1st order rate law to derive your expression for the velocity?

 

[dRic2]

Your book equation looks wrong dimensionally. Check their definition of the various k's.
Can you define the terms you use? Given those definitions, can you use the 1st order rate law to derive your expression for the velocity?

Which one are you referring to? because $R=R_{right}(1− \frac K K_{eq})$ has to be true. Anyway:
$C_i$ is the concentration of the species i (i.e. [mol/L]
$k$ is kinetic constant (i.e $R = k*C_i$) (it's dimension has to agree with
$C_i$
$K_{eq}$ is the constant of equilibrium $K_{eq} = exp(-\frac {ΔG_{R}} {RT})$ (dimensionless)
while $K = ∏a^{v_i}$ (which is equal to $K_{eq}$ when the equilibrium is reached) (dimensionless)

Regarding the second equation, I didn't notice it. I'll ask my professor, thank you

 

[dRic2]

Edit: Problem solved. And the second formula (corrected) is: $R = R_{right} - R_{left} = k(C_A - \frac {C_B} {K_{eq}} )$