Reaction rates constant and equilibrium constant

[kfcman289]

My Kaplan book told me that the equilibrium constant equals the forward reaction rate constant divided by the reverse reaction rate constant. The book also told me that reaction rates are affected by concentrations and enzymes. So how is it that the equilibrium constant is only affected by temperature and not concentrations?

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[Cmdr_Shepard]

The rate constants and the equilibrium constants are two different concepts.

Equilibrium constants are always K(eq) = [Products] / [Reactants]. This is only affected by temperature, because if you add reagent it will be converted to product and keep the ratio constant.

Rate constants are generally experimentally determined and depend on stoichiometry and conversion rates of reactant moles to product moles.

Edit: You can play with the algebra and divide rate laws to eventually find the equilibrium constant though. Is that what you mean?

 

[RocketSurgeon]

Think of it as total energy in the system heating or cooling the system changes the energy in the system so the product will energetically align itself to the energy in the system. More energy = more high energy product; Less energy = more lower energy reactant (assuming product here is higher energy)

Enzymes on the other hand don't add or remove energy but merely aid the energy transition process. So reactant can easily change into product and vice versa. They help reactants and products reach that energetically favorable equilibrium sooner without affecting it since they do not affect the energy of the system.

Hope that helped!

 

[Schenker]

My Kaplan book told me that the equilibrium constant equals the forward reaction rate constant divided by the reverse reaction rate constant. The book also told me that reaction rates are affected by concentrations and enzymes. So how is it that the equilibrium constant is only affected by temperature and not concentrations?

This is a pretty apt question. The answer is that factors which equally affect forward and reverse reaction rate constants won't affect the equilibrium constant.

Enzymes, for example, will increase the forward (kf) and reverse (kr) reaction rate constants by an equal factor (name this factor w), so Keq = (w*kf)/(w*kr) = kf/kr is unchanged.

Increasing the temperature, as Cmdr_Shepard corrected me below, will speed up both endothermic and exothermic reactions but will speed up endothermic reactions more than it speeds up exothermic reactions. Let's say the forward reaction is endothermic; this necessitates that the reverse reaction is exothermic. Thus, while kf and kr both increase when the temperature is raised, kf is increased by a greater factor than kr, meaning that Keq increases.

Finally, changing the concentrations of reactants/product does change the reaction rates but does not change the reaction rate constants. This is a good time to remember that the reaction rate and reaction rate constant are not the same thing; for the reaction A + B -> C the forward reaction rate (rf) is equal to kf[A][ B]. While rf will increase by a factor of 2 if the concentration of A is doubled, the reaction rate constant (kf) does not depend on the concentration of reactant. Since (unlike rf and rr) kf and kr are unaffected by changes in the concentration of reactants/produtcts, Keq is unaffected, since Keq = kr/kf. Remember, Keq is not equal to rr/rf.

*In case you are interested in why increasing the temperature increases kf by a greater factor than kr (assuming that the forward reaction is endothermic and thus that the reverse reaction is exothermic), it's because an endothermic reaction has a greater activation energy than its reverse reaction. Draw out an energy vs reaction coordinate diagram, and it should be pretty clear. I worked out the math using the Arrhenius equation, and increasing the temperature from T1 to T2 increases the reaction rate constant by a factor of exp(Ea*(1 - 1/x)*L) where Ea is the activation energy, x is the ratio of T2 to T1, and L = 1/(R*T1). It should be no surprise that this factor is dependent on temperature, but note that it is also dependent on activation energy, because that explains why an endothermic reaction speeds up more than its reverse reaction when the temperature is raised.

 

[kfcman289]

Thank you that makes so much sense now

 

[kfcman289]

One more question. The book says that termperature will increase the reaction rate because particles are colldiing faster more with more energy, but you are saying certain reaction rates will decrease due to an increase in temeperature?

 

[Cmdr_Shepard]

All reaction rates will increase with an increase in temperature. But the equilibrium will shift to offset the increase in temperature. To reactants for exothermic, and to products for endothermic ones.

 

[Schenker]

Ah, I confused reaction rates and reaction rate constants and may have misled you in my post. I've corrected it.

Also, @Cmdr_Shepard and your book are right about all reaction rates increasing with an increase in temperature.