[jennielynn_1980]

I am having difficulty understand how to write a rate law and also what a rate constant represents.

My first question is:
What would be the effect of an increase in temperature on the rate constant of the forward reaction and the reverse reaction?

The second question is:
Does anyone know of an online tutorial or webpage about writing rate laws because I can't do it to save my life. 

Thanks.

[sdekivit]

do you know what happens with any reaction when the temperature increases ? and will the increase in temperature have more effect on an endothermic or exothermic reaction ?

when speaking of a forward and backward reaction, we speak of an equilibrium. So you have to determine what reaction is exothermic and what reaction is endothermic to decide what happens with the reactions involvend in an equilibrium.

[jennielynn_1980]

when the temperature increases, the rate of reaction increases.  An increase in temperature has more of an effect on endothermic reactions because it will change the activation energy more for an endothermic reaction.  I am assuming a forward reaction is exothermic and the reverse in endothermic. 

I think I found the answer:
for the forward reaction the value of the rate constant would become smaller and the for the reverse reaction the value of the rate constant would become larger.

Is that correct?

[Donaldson Tan]

when the temperature increases, the rate of reaction increases. 

The rate constants for both backward and forward reactions increases with temperature.

An increase in temperature has more of an effect on endothermic reactions because it will change the activation energy more for an endothermic reaction.

You are indirectly rephrasing Le Chatelier's Principle on the effect of heat on chemical equilibrium.

The equilibrium constant is temperature-dependent constant. The ratio of the forward rate constant to the backward rate constant depends on the final temperature. In general, dKc/dT is positive increasing if the forward reaction is endothermic.

[jennielynn_1980]

Why does the rate constant increase for both?
Is the rate constant the same as the equilibrium constant?

[jennielynn_1980]

Okay, I think I get it now.  The rate constant increases for both because the temperature would increase the reaction rate for both.

Do I have it now?

[sdekivit]

Assume the reaction A <--> B

with reaction rate s for the reaction A --> B: s = k₁ * [A]
with reaction rate s for the reaction B --> A: s = k_-1 *

Let A --> B be an endothermic reaction. Now because there is equilibrium the following is valid:

k₁ * [A] =  k_-1 * and thus /[A] = k₁ / k_-1 = K_ev

Now we let T increase and we already know that the reaction rate constant k for an endothermic reaction increases moere then that of a exothermic reaction.

--> when k₁ for the endothermic reaction increases more then k_-1 of the exothermic reaction, k₁ / k_-1 rises and thus increasing the equilibrium constant K_ev resulting in a shift in the equilibrium to the right side / forward direction (because /[A] has risen [A] is decreased and is increased).

( note that the rise in equilibrium constant is characterisic for a rise in temperature. )

[Donaldson Tan]

Do I have it now?

Yes. I also want to mention that component proportion does not vary linearly with reaction rate.

[jennielynn_1980]

Thanks to you both 

[jennielynn_1980]

So I understand rate laws a little better now   BUt I have another question regarding how to write a rate law.
The info I have is this from a simulated experiment:
if you double the concentration of iodide ions, the reaction time is cut in half
if you doublee the concentration of persulphate ions, the reaction time is cut in half

So, this means the reaction rate doubles right?  I think I am confused with regards to the rate of reaction doubling or tripling and so forth with the actual time.  Like does the rate of reaction increase if the amount of time it takes for the reaction to happen is less? 

So assume I think I know what I am doing the rate law would be:

rate = k [I^-]²[S₂O₈^2-]²

[Donaldson Tan]

So assume I think I know what I am doing the rate law would be:

rate = k [I^-]²[S₂O₈^2-]²

If you plugin test values into your equation, you would find that your equation predicts the reaction rate to quadraple if the concentration of either reactant is doubled. However, your reaction rate should only double if the concentration of either reactant doubles, so the reaction should be first order to either reactant.

[jennielynn_1980]

I don't understand what this means "your reaction should be first order to either reactant"

Does this mean that there should be no exponents?  I think you mean it should look like this:

rate = k[I^- ][S2O8^2-]

[Donaldson Tan]

I think you mean it should look like this:

rate = k[I^- ][S2O8^2-]

YES

[jennielynn_1980]

Thanks