Arrhenius equation

Dear roes,
Arrhenius equation is of the form
k=A*e(-Ea/RT)
(read e as exponential to)
Where Ea is the activation energy which is always possitive.
For an exothermic reactio, the exponent in the Arrhenius equation is negative.
dk/dt = (A*Ea/ R*Tsquare)* e(-Ea/R*T) is greater than zero;
So k increases as temperature increases.
As temperature decreases, the rate of reaction decreases so the conversion decreases.

Atul.

roes, See the following site:
It will explain everything you ever wanted to know and then some.

Hope this helps.
saxon

I'll try to explain the effects of temperature on the rates of a reaction and on the reaction equilibrium meaning the concentration of reaction products (or as roes says, conversion).

Chemical rate laws are those that try to relate the rates of a reaction (i.e., changes of concentration per unit time) as function of concentrations and of thermodynamic parameters. The Arrhenius equation is one rate law trying to correlate rates with temperatures.

In general, to better understand the effect of the various functions in an exponential equation one takes logs.
For example, using atul1's notation:

lnk = lnA - E_a/RT

where A, E_a , R are all constant parameters.

Not all reactions follow the Arrhenius formula, meaning that in a plot of lnk vs 1/T you don't always get a straight line. However, on short T ranges you may get a straight line, and can thus calculate the activation energy E_a. Activation energy is always negative, meaning one has to add it to the reactants for them to start the reaction.

A general explanation of why higher temperatures increase the rates of a reaction is based on the kinetic theory that implies that at higher temperatures the rate of collisions between "reacting" molecules increases.

As to predicting the effect of temperature on how the reaction equilibrium (not the rate) will respond to it, Le Chatelier principle comes to our rescue, stating that if a reaction is exothermic, lowering the temperature will favour the increase in the amount of the reaction products [conversion], because the heat generated in the reaction will tend to minimize the lowering of temperature. For example, take the exothermic production of ammonia by the Haber process.

Now, combining activation energy with temperature: the higher the activation energy of a reaction, the more sensitive is its rate k to changes of temperature. The activation energy E_a is larger for endothermic reactions than for their reverse exothermic counterpart. Therefore, as the temperature rises the forward (endothermic) reaction produces more products and the rate increases more rapidly until the concentration of products has risen enough for the reverse reaction rate to match the forward rate.

Coming back to the Haber process. Because the synthesis of ammonia is exothermic, lower temperatures favour the product (better conversion). However, the rate at which nitrogen and hydrogen combine at ambient temperature is virtually zero and the reaction proceeds infinitely slowly towards equilibrium. Fritz Haber was faced with a dilemma: high temperatures would be needed to get an acceptable rate of conversion but if he did increase temperatures the extent of conversion would be very low. He solved the conumdrum by using a catalyst, porous iron, that reduced the activation energy and could thus synthesize ammonia at lower temperatures. However, the catalyst also sped up the reverse reaction, i.e., although the equilibrium composition was reached more quickly, it contained very little ammonia! The final solution was using Le Chatelier again, he increased the pressure (favouring lower volumes) and removed the ammonia as soon as it was formed. His chemical engineering colleague, Carl Bosch, successfully designed the first high-temperature, high-pressure, catalytic industrial process, that led to two Nobel prizes. The Haber-Bosch process is still the sole source of ammonia produced throughout the world.

Summarizing, higher temperatures increase rates of either endothermic or exothermic reactions, but do not induce conversion of reactants to products (i.e., equilibrium) in exothermic reactions.

I hope my above expose is of help.

roes,
see the excellent site given by saxon, and also view the ‘Kinetics’ and ‘Equilibria’ links on the LHS of the page. Then, you can understand the following: Rate kinetics and equilibrium conversions are not rigorously linked and are often inversely related.

There seems to be some confusion between kinetics and equilibria concepts in your question and in Atul’s answer. For all positive activation energies (normal), whether an endothermic or exothermic reaction, the reaction rate increases with temperature. Exothermic reactions sometimes generate enough heat that they self-heat and autoaccelerate (self-propagate).

So ‘does conversion increase or decrease as temperature in increased for an exothermic reaction?’ An example is given where conversions vary as follows:
“If you increase the temperature, then the endothermic reaction will be favored because that will take in some of the excess heat. If you decrease the temperature, the exothermic reaction will be favored because it will produce the heat that was lost.”
at Atul’s answer is only correct if there is insufficient reaction time to reach equilibrium. The yield is normally determined by (or limited by) the reaction equilibrium constant K, which is a function of &[ignore]Delta[/ignore];G_Reaction, which also includes entropy. See

25362,
Very good answer. Like your Haber cycle example.
I will add that Arrhenius won the 1903 Nobel Prize for chemistry.

Kenvlach, I should have said the activation energy term is always negative, meaning one has to add it to the reactants ...
Thanks for your appreciation.

Hold On.....

Fellows... DO NOT CONFUSE THERMODYNAMICS with KINETICS
Which most you have Done!!!. AGAIN DO NOT CONFUSE.

ENDO or EXO-thermicity is a THERMODYNAMIC PROPERTY
for the species involved, and it depends on the HEAT of FORMATION (say Delta H = DH_p - DH_r, I will not explain it here, if you need I will do later on) of the reactant and product. Furthermore, Delta H comes with a NEGATIVE sign.
that means -Delta H = (DH_p -Dh_r). Are you following?
Then from DH_p-DH_r, you get a numerical value with a sign.
For EXOTHERMIC REACTION, -Delta H (note I am always talking -Delta H !) is NEGATIVE, where as -Delta H is POSITIVE for ENDOTHERMIC REACTION

I will take 25362's example of NH3 process:
0.5 N2 + 1.5 H2 = NH3.........(1)
At some T
DH N2 = x
DH H2 = y
DH NH3 = z

For the reaction (1), -Delta H = z-(x+y) .......(1a)
For NH3, the RHS term of Eq (1a)z-(x+y) is ALWAYS ALWAYS Negative. This is EXOTHERMIC REACTION.

I will an example of Steam methane reformer (use in any NH3 process).

CH4 + H2O = CO + 3 H2-----------(2)
At some T the heat of formation for the sepcies involved
DH CH4 = d
DH H2O = e
DH CO = f
DH H2 = g
So, for the reaction

-Delta H = (f+3g)-(d+e)------(2a)
Do this calculation and you will find, the NUMERICAL VALUE of RIGHT HAND SIDE of Eq. (2a) is POSITIVE

So WHAT??

HEAT of FORMATION NEGATIVE MEANS, the reaction produces HEAT after the products are formed. So, with TEMPERATURE, (the tendency of MOTHER NATURE) the THE THERMODYNAMIC CONVERSION WILL DECREASE.

NOTE, THERMODYNAMICS DOES NOT SAY HOW FAST WE WILL REACH THE PRODUCT. THAT's WHERE KINETICS comes into PLAY!!!
Reaction Kinetics itself is a big subject. i will refrain for the discussion. Only I would like to say about k=A*Exp(-E/RT)
REMEMBER, ALWAYS ALWAYS ALWAYS (I can not stress any more)
E- the activation energy is ALWASY POSITIVE, EVEN if you go to Heaven or Hell- IT IS POSITIVE.BUT ALSO REMEMBER it is PRE-FIXED with NEGATIVE SIGN. So what do we get out of this?
It says, RATE OF REACTION WILL ALWAYS INCREASE WITH TEMPERATURE...................

CONFUSED AGAIN? Your Question: But why do we see conversion drop with higher temperature? THIS IS BECAUSE On the hireachery, THERMODYNAMICS>KINETICS. Thermodynamics determines the MAXIMUM POSSIBLE CONVERSION at a GIVEN CONDITION, via Equilibrium constant, Gibbes Free Energy = -RT LN(K_Eq). Thermodynamics is the UPPER LIMIT.

So, FEW THINGS YOU NEED TO KNOW:

1. Is the REACTION AT ALL POSSIBLE?
Ans: FreeEnergy (say Gibbs Free energy) of the System (i.e., reactant and product) will tell you that. Remember DG= -RT lnK_eq)

2. Endo or Exothermic?
Ans: Heat of Formation (or PRD & Reactant) will tell you that

For most of the compound those two are available. what is not available (or easily available) is the HOW FAST WE CAN GO THERE? THE RATE.. and that is KINETICS

3. rate= f(T) * f(conccentration)
where f is the function. And f(T) goes back to Arrhenius type expression. f(T) = k = A Exp(-E/RT). Also remember to quote both A, and E for a reaction (A-E Paired), and E is ALWAYS POSITIVE. A quick example, put a -ve value, say X
k= A Exp{-(-X/RT)} = A Exp{X/RT}... Hey.. RUN AWAY Exponential of a +VE NUMBER ... use proper unit to see the value........TILL THEN BYE

Jay

Jay,
a) No need to shout.

b) Please use subscripts and Greek letters ( for better readability.

c) IMHO, you haven’t added much to 25362's or my explanations, and in any case, the links in saxon’s and my post give very clear answers. In one remark by 25362, yield would have been better explained in terms of Gibbs energy instead of endo/exo, but there is a statistical thermodynamics basis to his answer.

d) Why re-state 25362’s comment that NH₃ formation is exothermic?
The important point regarding development of the Haber cycle and yield: &[ignore]Delta[/ignore];^oG_f,NH3 changes sign at about 464^oK, becoming + at higher temperatures, unfavorable for yield, hence necessitating use of Le Chatelier’s Principle (in this case, higher P) to push the reaction:

&[ignore]Delta[/ignore];^oG_f,NH3 = &[ignore]Delta[/ignore];^oH_Reaction – T &[ignore]Delta[/ignore];^oS_Reaction

= -45900 + 99.05T J/mole
{Note: used 298 K values from CRC handbook. To be rigorous, &[ignore]Delta[/ignore];H and &[ignore]Delta[/ignore];S are f(T)’s.}

All in all, though, I welcome your contribution. Haven’t encountered someone so passionate about thermodynamics in quite a while.

to jbando, concerning reaction kinetics, and in addition to what kenvlach says on reaction thermodynamics, may I say life is not so simple. In your point 3 there is no statement concerning the fact that rates of reaction are also influenced by (are a function of) two more factors -besides temperature and concentration- the presence of (homogeneous and/or heterogeneous) catalysts, and the available surface area of the reactants in heterogeneous reactions. These may act by affecting A or E, in the Arrhenius equation, however, merit to be separately mentioned.

One word about multiple reactions: the rate-determining step, i.e., the one that governs the rate of the overall reaction, is the slowest. Then one shouldn't forget chain reactions in which one intermediate reacts to produce another. In radical chain reactions the intermediates or chain carriers are radicals. These reactions may include characteristics such as initiation, propagation, retardation, inhibition and termination steps, as well as chain branching that, in some cases, may lead to explosions.

It would be superfluous to mention the wide variety of reaction mechanisms. Wouldn't it ?