Gen Chem #77 (Destroyer) Activation Energy with Temperature

[dentalphobia]

DAT Destroyer Gen Chem # 77

Which of the following will lower the activation energy ?

a) using a catalyst
b) lowering the temperature
c) raising the temperature
d) increasing the pressure
e) increasing the volume

answer (a)
------------------------------------

I thought the answer can be (a) and (c) by the arrhenius equstion.

Ea=-RT ln (k/A)

However, wikipedia said the term T is cancelled by temperature decepdence 'k'.

http://en.wikipedia.org/wiki/Activation_energy

It's bit hard to understand this part.

Anybody has this problem ?

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[bennijai]

The activation energy will not be lowered by heat. Heat will only increase the rate of reactions. So heat will bring molecules closer to the activation energy but will not lower activation energy.

A catalyst is the only way to lower activation energy.

 

[sixkiller]

DAT Destroyer Gen Chem # 77

Which of the following will lower the activation energy ?

a) using a catalyst
b) lowering the temperature
c) raising the temperature
d) increasing the pressure
e) increasing the volume

answer (a)
------------------------------------

I thought the answer can be (a) and (c) by the arrhenius equstion.

Ea=-RT ln (k/A)

However, wikipedia said the term T is cancelled by temperature decepdence 'k'.

http://en.wikipedia.org/wiki/Activation_energy

It's bit hard to understand this part.

Anybody has this problem ?

k = Aexp(-Ea/RT)

A is the frequency of collisions with proper orientation

exp(-Ea/RT) is the frequency of collisions with sufficient kinetic energy

Thus, the rate constant k is the frequency of EFFECTIVE collisions; those with sufficient energy and proper orientation.

Thus, raising the T raises specifically, the frequency of collisions with sufficient kinetic energy, but this may also be interpreted as a raise in the frequency of effective collisions.

Activation energy for a given mechanism is constant as is the the A factor; neither depend on temperature.

 

[nze82]

Think of the activation energy of a reaction as the steepness of a hill. Think of the reactants as a group of people trying to climb this hill (complete a reaction). Think of heat as some energy drink that increases the energy of these people. If they drink it, they can climb the hill more easily (reaction takes place faster), but the steepness of the hill doesn't change upon consumption of this drink (activation energy remains the same).

 

[dentalphobia]

Thanks, folks.

However, my question was solely based on the Arrhenius equation below.

Ea=-RT ln (k/A)

Ea is Activation energy. Here increasing T, the value of Ea will be lower.

How could you explain this ?

 

[UCB05]

Thanks, folks.

However, my question was solely based on the Arrhenius equation below.

Ea=-RT ln (k/A)

Ea is Activation energy. Here increasing T, the value of Ea will be lower.

How could you explain this ?

Your interpretation of the equation is a bit off. The Arrhenius equation approximates the relationship between T and the rate constant k. If T increases, k increases. R is a universal constant and Ea and A are constants determined by the reaction and do not change at all.

If that does not make sense, your logic is analogous to saying that in PE=mgh, that if you increase PE, you increase mass. For the particular system, you assume m is constant. Sure you can change the mass, but then you're no longer talking about the same object, just as to change Ea in the Arrhenius equation is to change the nature of the reaction.

If you fiddle around with the equation, you can get R = -Ea/T ln(k/A). Does that mean that increasing T decreases R? Of course not. For a particular reaction, R, A, and Ea are constants. The variables are T and k

 

[sixkiller]

It also helps to realize that the arrhenius equation is an empirical equation....

Like UCB pointed out regarding PE....also Newton's second law comes to mind...

 

[Shinpe]

Another example is Resistance = Voltage/current. If you increase the voltage across a resistance, does the actual resistance increase? No, Just the current also increases accordingly to keep R constant.

Here, if you increase T, K is increasing accordingly to keep Ea constant.