rate constant with increase in temp& presence of catalyst

[inaccensa]

Plz Correct me if I am wrong

With increasing temp, the rate of the reaction increases, however both the forward and the reverse reaction rate will also increase. The rate constant K increases with increasing temp & temp doesn't affect the energy of activation.

Catalyst on the other hand, do not affect the rate constant K. They increase the rate of reaction, which again is both the forward and the reverse rates? It also decreases the energy of activation

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[dragon529]

I might not agree with you that temperature alone won't affect Energy of Activation... since a higher temperature allows a better chance for molecules to collide...

 

[inaccensa]

I might not agree with you that temperature alone won't affect Energy of Activation... since a higher temperature allows a better chance for molecules to collide...

actually EK says that temp does affect the activation energy, but it is small enuf to be ignored for MCAT. i'm just trying to get the general idea if what i understand about each is correct

 

[sleepy425]

Plz Correct me if I am wrong

With increasing temp, the rate of the reaction increases, however both the forward and the reverse reaction rate will also increase. The rate constant K increases with increasing temp & temp doesn't affect the energy of activation.

Catalyst on the other hand, do not affect the rate constant K. They increase the rate of reaction, which again is both the forward and the reverse rates? It also decreases the energy of activation

The first part is correct. Increasing temp increases both rates. You're also correct that the temperature does not affect the energy of activation, since the energy of activation is a fundamental property of the molecules and transition states involved. Increased temperature increases the energy of the molecules, which makes it easier for the molecules to get over the energy barrier. But the barrier is still just as high.

There is one part of that first paragraph that I have a problem with. You keep referring to "the rate constant K." But in a reversible chemical reaction, there are two rate constants. There is the rate for the forward direction, and the rate for the reverse direction. In the case of increasing temperature, both the forward and reverse rate constants increase. Perhaps you're referring to the equilibrium constant, which is the ratio of the forward and reverse rate constants. Equilibrium constants are temperature dependent, and the temperature dependence is given by the Van't Hoff equation (basically, if the change in enthalpy of the reaction is negative, then the equilibrium constant decreases with increasing temp, if it is positive, the equilibrium constant increases with increasing temp).

Your second paragraph is partially incorrect. Catalysts increase both the forward and the reverse rate constants (this is known as the principle of microscopic reversibility). This is how a catalyst increases the rate of a chemical reaction. However, catalysts do not affect the equilibrium constant.

Incidentally, if you're not sure if something affects the rate constant of a chemical reaction, but you know that the rate is affected, here's how you can tell if the rate constant is affected:

The rate of a chemical reaction is given as

Rate=k[A]^m*^n...

where A and B are reactants m and n are the respective orders of those reactants.

So, the rate only depends mathematically on two things, the rate constant and the concentration of reactants. Soooo, if you know the rate is affected, and you know the concentration of reactants is not affected, then the rate constant has to be affected.

 

[inaccensa]

The first part is correct. Increasing temp increases both rates. You're also correct that the temperature does not affect the energy of activation, since the energy of activation is a fundamental property of the molecules and transition states involved. Increased temperature increases the energy of the molecules, which makes it easier for the molecules to get over the energy barrier. But the barrier is still just as high.

There is one part of that first paragraph that I have a problem with. You keep referring to "the rate constant K." But in a reversible chemical reaction, there are two rate constants. There is the rate for the forward direction, and the rate for the reverse direction. In the case of increasing temperature, both the forward and reverse rate constants increase. Perhaps you're referring to the equilibrium constant, which is the ratio of the forward and reverse rate constants. Equilibrium constants are temperature dependent, and the temperature dependence is given by the Van't Hoff equation (basically, if the change in enthalpy of the reaction is negative, then the equilibrium constant decreases with increasing temp, if it is positive, the equilibrium constant increases with increasing temp).

Your second paragraph is partially incorrect. Catalysts increase both the forward and the reverse rate constants (this is known as the principle of microscopic reversibility). This is how a catalyst increases the rate of a chemical reaction. However, catalysts do not affect the equilibrium constant.

Incidentally, if you're not sure if something affects the rate constant of a chemical reaction, but you know that the rate is affected, here's how you can tell if the rate constant is affected:

The rate of a chemical reaction is given as

Rate=k[A]^m*^n...

where A and B are reactants m and n are the respective orders of those reactants.

So, the rate only depends mathematically on two things, the rate constant and the concentration of reactants. Soooo, if you know the rate is affected, and you know the concentration of reactants is not affected, then the rate constant has to be affected.

That explains a lot. Thank you very much

 

[Hemichordate]

Is this correct:

Rate constant depends on: temperature, pressure, catalysts
Rate depends on: temperature, concentration of reactions/products, catalysts

Anything else?

 

[stockraider]

the rule is rate constant (k)= A*e^-(activation energy/RT), where A is the arhenius factor that is unique to each reaction, but beyond the scope of the mcat. I have one problem with your statement...enzymes do effect the activation energy as can be seen with this equation since enzymes lower the activation energy, they will increase k, if you do the math you will see it in the equation. also in regards to temperature...if you increase temp, as seen in the equation, you will increase k as well. so rule of thumb is decreasing activation energy or increasing temp BOTH increase the rate constant. just memorize the above equation and you'll be fine. another way to write it is ln k= ln A- (activation energy/RT). this equation is on pg. 547 of the 2008 TPR physical sciences review book if you want a reference for it and a more elaborate explanation.

 

[sleepy425]

Rate constant depends on: temperature, pressure, catalysts
Rate depends on: temperature, concentration of reactions/products, catalysts, pressure

It's correct, but I just added in pressure to your bottom list because if the rate constant depends on pressure, then the rate itself will also depend on pressure. I can't think of anything else that might significantly affect them.

 

[stockraider]

Is this correct:

Rate constant depends on: temperature, pressure, catalysts
Rate depends on: temperature, concentration of reactions/products, catalysts

Anything else?

rate=k*[A]*, so temp effects it indirectly by directly effecting k and obviously the concentrations effect it, and the catalyst indirectly effects it by effecting k.
the rate constant, k is effected by the variables in the equation rate constant (k)= A*e^-(activation energy/RT)...see my previous post for that.

 

[stockraider]

It's correct, but I just added in pressure to your bottom list because if the rate constant depends on pressure, then the rate itself will also depend on pressure. I can't think of anything else that might significantly affect them.

really quickly can someone explain to me how pressure effects the rate constant?

 

[sleepy425]

I can only explain it empirically, since the mathematics behind it is a bit complicated. If the reagents are solid or liquid, pressure has little effect (there is an effect, but it is minimal since solids and liquids are only marginally compressible). For reactions between gases, increased pressure increases the rate constant since the pressure increases the rate of molecular collisions.

 

[stockraider]

I can only explain it empirically, since the mathematics behind it is a bit complicated. If the reagents are solid or liquid, pressure has little effect (there is an effect, but it is minimal since solids and liquids are only marginally compressible). For reactions between gases, increased pressure increases the rate constant since the pressure increases the rate of molecular collisions.

ah gotcha, so for gas this is true, and it makes sense increasing the pressure in a fixed container means increasing the subsequent temperature (using the combined gas law), meaning the rate constant in the above equation is affected...thanks

 

[inaccensa]

The first part is correct. Increasing temp increases both rates. You're also correct that the temperature does not affect the energy of activation, since the energy of activation is a fundamental property of the molecules and transition states involved. Increased temperature increases the energy of the molecules, which makes it easier for the molecules to get over the energy barrier. But the barrier is still just as high.

There is one part of that first paragraph that I have a problem with. You keep referring to "the rate constant K." But in a reversible chemical reaction, there are two rate constants. There is the rate for the forward direction, and the rate for the reverse direction. In the case of increasing temperature, both the forward and reverse rate constants increase. Perhaps you're referring to the equilibrium constant, which is the ratio of the forward and reverse rate constants. Equilibrium constants are temperature dependent, and the temperature dependence is given by the Van't Hoff equation (basically, if the change in enthalpy of the reaction is negative, then the equilibrium constant decreases with increasing temp, if it is positive, the equilibrium constant increases with increasing temp).

Your second paragraph is partially incorrect. Catalysts increase both the forward and the reverse rate constants (this is known as the principle of microscopic reversibility). This is how a catalyst increases the rate of a chemical reaction. However, catalysts do not affect the equilibrium constant.

Incidentally, if you're not sure if something affects the rate constant of a chemical reaction, but you know that the rate is affected, here's how you can tell if the rate constant is affected:

The rate of a chemical reaction is given as

Rate=k[A]^m*^n...

where A and B are reactants m and n are the respective orders of those reactants.

So, the rate only depends mathematically on two things, the rate constant and the concentration of reactants. Soooo, if you know the rate is affected, and you know the concentration of reactants is not affected, then the rate constant has to be affected.

So the take home lesson is as follows
-temperature doesnot affect the activation energy, but rather enables more reactants to reach activation energy quickly.

-The rate of reaction increases with increasing temperature, presence of catalyst, increasing the conc of reactants and also pressure(for gases). It will increase the rate of both forward and reverse reactions and the net change is zero.

-The rate constants are affected by both temp and catalysts. Also acc to another post pressure will also affect the rate constant. Similarly both the forward and the reverse rate constants are affected and the net change is zero

-The equilibrium constant is dependent on temperature. Can you explain the relationship between K and temp. You keep referring to Enthalpy, so is there a formula based on that relationship?

- I keep getting confused with natural logs. Acc to the formula, it looks like ln K=ln A - Ea/RT, K is inversely proportional to temp. I know I'm probably messing up on the logs. Can you please kindly explain that

 

[stockraider]

So the take home lesson is as follows
-temperature doesnot affect the activation energy, but rather enables more reactants to reach activation energy quickly.

-The rate of reaction increases with increasing temperature, presence of catalyst, increasing the conc of reactants and also pressure(for gases). It will increase the rate of both forward and reverse reactions and the net change is zero.

-The rate constants are affected by both temp and catalysts. Also acc to another post pressure will also affect the rate constant. Similarly both the forward and the reverse rate constants are affected and the net change is zero

-The equilibrium constant is dependent on temperature. Can you explain the relationship between K and temp. You keep referring to Enthalpy, so is there a formula based on that relationship?

- I keep getting confused with natural logs. Acc to the formula, it looks like ln K=ln A - Ea/RT, K is inversely proportional to temp. I know I'm probably messing up on the logs. Can you please kindly explain that

in regards to the equation ln k= ln A- Ea/RT, k is NOT inversely proportional to T, it is directly proportional to it. think about it, increasing T will decrease the overall term Ea/RT, which if it's being subtracted from ln A, will take away LESS from ln A, making ln k, and thus k larger. i know this is a lot to follow, but just realize increasing T will lower the term being subtracted, which will increase the overall sum of what ln k equals.

 

[inaccensa]

in regards to the equation ln k= ln A- Ea/RT, k is NOT inversely proportional to T, it is directly proportional to it. think about it, increasing T will decrease the overall term Ea/RT, which if it's being subtracted from ln A, will take away LESS from ln A, making ln k, and thus k larger. i know this is a lot to follow, but just realize increasing T will lower the term being subtracted, which will increase the overall sum of what ln k equals.

That makes sense. Thanks.

Decreasing Ea will increase K
Increasing T will increase K
Since R is constant, we can ignore that.

So k here is the rate constant and not equilibrium constant. Equilibrium constant is only affected by temp. If the conc of reactant and products change, K will be affected, correct?

 

[inaccensa]

So the take home lesson is as follows
-temperature doesnot affect the activation energy, but rather enables more reactants to reach activation energy quickly.

-The rate of reaction increases with increasing temperature, presence of catalyst, increasing the conc of reactants and also pressure(for gases). It will increase the rate of both forward and reverse reactions and the net change is zero.

-The rate constants are affected by both temp and catalysts. Also acc to another post pressure will also affect the rate constant. Similarly both the forward and the reverse rate constants are affected and the net change is zero

-The equilibrium constant is dependent on temperature. Can you explain the relationship between K and temp. You keep referring to Enthalpy, so is there a formula based on that relationship?

- I keep getting confused with natural logs. Acc to the formula, it looks like ln K=ln A - Ea/RT, K is inversely proportional to temp. I know I'm probably messing up on the logs. Can you please kindly explain that

So a q in EK states that catalysts increase the rate of reaction, I do agree with that statement, but the net change is zero, since it increases the rate of forward and reverse. When they ask such a q, are they talking about how the actual rate of both forward and reverse increase in real terms?