Chemical Kinetics: velocity and equilibrium

In summary, the conversation is discussing the velocity of a chemical reaction and how it is affected by approaching equilibrium. One person mentions a formula they have come up with to take equilibrium into account, but they are unsure if it is the only one. The other person questions the validity of a formula from a book, and the first person agrees that it looks dimensionally wrong. They go on to define the terms used in their equations and discuss a correction to the second formula.
  • #1
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
$$R = R_{right} - R_{left} = R_{right}(1 - \frac K K_{eq}) = kC_A(1- \frac K K_{eq} ) $$ ?
 
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  • #2
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?
 
  • #3
mjc123 said:
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
 
  • #4
Edit: Problem solved. And the second formula (corrected) is: ##R = R_{right} - R_{left} = k(C_A - \frac {C_B} {K_{eq}} )##
 

1. What is chemical kinetics?

Chemical kinetics is the study of the rates of chemical reactions and the factors that affect them. It involves analyzing the speed at which reactants are converted into products and understanding the mechanisms of these reactions.

2. What is meant by reaction velocity?

Reaction velocity, also known as reaction rate, is the speed at which a chemical reaction takes place. It is typically measured in terms of the change in concentration of either the reactants or products over time.

3. What factors affect reaction velocity?

Several factors can affect reaction velocity, including temperature, concentration of reactants, presence of a catalyst, and surface area of reactants. These factors can either increase or decrease the rate of a chemical reaction.

4. What is chemical equilibrium?

Chemical equilibrium is a state in which the rates of the forward and reverse reactions in a chemical system are equal. This means that the concentrations of reactants and products remain constant over time, although the reactions are still occurring.

5. How is equilibrium affected by changes in temperature and concentration?

Temperature changes can affect equilibrium by shifting the equilibrium position towards the endothermic (heat-absorbing) or exothermic (heat-releasing) direction. Changes in concentration can also affect equilibrium by changing the ratio of reactants to products, which can shift the equilibrium position in favor of the reactants or products.

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