Temperature and Le Chatelier's principle

[TheMightyBoosh]

While reading about equillibrium in TBR general chemistry I understood that temperature will act as a stress to shift a reaction in a certain direction (heating system will result in shift towards endothermic direction). However, in EK 1001 Chemistry problem 289, when asked what effect temperature will have on the forward rate for an exothermic reaction (2CO(g) + O2 (g) -> 2CO2 (g), when starting with only reactants), they stated that the forward reaction rate will increase because raising the temperature always increases the rate of of the reaction.

Can someone explain this discrepancy?

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[loveoforganic]

Temperature ALWAYS increases rate because it increases the percentage of molecules above the threshhold energy value to react.

LeChatlier's principle states that a closed reaction environment will equilibrate in such a way as to minimize stress. The equilibrium maintained could be at near 100% reactants at high temperature, but regardless, the forward and reverse reactions will still be very quick. The forward reaction would just be much slower than the reverse reaction.

 

[TheMightyBoosh]

I think I get you.

So an increase in temperature will affect the kinetics of the reaction by increasing the forward and reverse rate towards equilibrium, while at the same time shifting the equilibrium towards the reactant side... am I right? And if I understand correctly, this is different than other stresses (changing pressure, volume, moles) in that the Keq values for reactions involving those other stresses will not change?

 

[loveoforganic]

I believe that everything you've said is correct (we'll say 90% sure) . So if someone else wants to verify, that'd be great.

 

[TheMightyBoosh]

I believe that everything you've said is correct (we'll say 90% sure) . So if someone else wants to verify, that'd be great.

haha 90% is enough for me to go to bed Thanks for the quick response!

 

[Fifty 3rds]

TheMightyBoosh,

I believe what you said is correct. I believe it is safe to assume that anytime temperature is raised, the K value will raise (products over reactants) although BR had a question that emphasized a different thought. I'm not near my book but I believe it had a double displacement reaction with more moles of gas being on the right side than on the left.

But I don't quite remember it all. I will reply back when I get home.

 

[Phantastic]

The key here is that all reactions, even exothermic ones, require some heat to get going (activation energy of transition state). Whether or not they let out heat afterward depends on how stable the product formed is.

Therefore, increasing T should always increase the rate by increasing little k (the rate constant). But yes, that will increase the rate of the forward and reverse reactions, so Keq should be unchanged.

"The only way to change K is to change T" is an accurate statement, but there is a caveat. Exothermic reactions and endothermic reactions will display different thermodynamic shifts in equilibrium to the same temperature stimulus. e.g. An endothermic reaction creates more products (K gets bigger) when I add heat (increase T). This is because heat can be considered a reactant, and adding reactant will shift the equilibrium to the product side.

So, here we have an exothermic reaction. If we raise the temperature, it is like adding products to the system. Yes, we increase the rate of the forward and reverse reaction by increasing T, but this keeps K the same...if we add heat (a product), we will necessarily be shifting the reaction to the reactants side (K gets smaller).

I think EK's explanation is incorrect.

 

[Fifty 3rds]

Once again, great reply Phantastic.

 

[TheMightyBoosh]

The key here is that all reactions, even exothermic ones, require some heat to get going (activation energy of transition state). Whether or not they let out heat afterward depends on how stable the product formed is.

Therefore, increasing T should always increase the rate by increasing little k (the rate constant). But yes, that will increase the rate of the forward and reverse reactions, so Keq should be unchanged.

"The only way to change K is to change T" is an accurate statement, but there is a caveat. Exothermic reactions and endothermic reactions will display different thermodynamic shifts in equilibrium to the same temperature stimulus. e.g. An endothermic reaction creates more products (K gets bigger) when I add heat (increase T). This is because heat can be considered a reactant, and adding reactant will shift the equilibrium to the product side.

So, here we have an exothermic reaction. If we raise the temperature, it is like adding products to the system. Yes, we increase the rate of the forward and reverse reaction by increasing T, but this keeps K the same...if we add heat (a product), we will necessarily be shifting the reaction to the reactants side (K gets smaller).

I think EK's explanation is incorrect.

So just to clarify your explanation, do you agree with lovefororganic's explanation about how in an exothermic reaction, adding heat should cause the reverse rate of the reaction to increase more than it will cause the forward reaction rate to increase? Essential, while little k forward and backward get bigger, the ratio of little k forward/little k backward should be smaller than it was before heat was added, making big K smaller.

I hope I have this right.

 

[eablackwell]

Heating it will increase the rate, period. I think you are overthinking the problem.

Did the problem state that the system included this particular reaction at equilibrium, and then you added heat? Or did it just say here's a reaction, and some heat...what happens?

If it didn't state it was a reversible reaction at equilibrium why even bother thinking in shifts? Shifts are to maintain the equilibrium state.

Think in kinetics. I think you're confusing 2 different ideas.

Just 2 cents from a chem teacher

 

[TheMightyBoosh]

in EK 1001 Chemistry problem 289, when asked what effect temperature will have on the forward rate for an exothermic reaction (2CO(g) + O2 (g) -> 2CO2 (g), when starting with only reactants), they stated that the forward reaction rate will increase because raising the temperature always increases the rate of of the reaction.

So the question asked whether the rate of the forward reaction would increase, decrease or stay the same. I understand that due to kinetics the reaction rate will increase in the forward direction. No further explanation needed here.

What I wanted to clarify was what would happen with the equilibrium position in this scenario. Will it move to the left? Or does temperature change keq only when the reaction is already at equillbrium?

And yes, I over think things... it's my favorite way to learn

 

[eablackwell]

If at equilibrium the reaction will shift left. Via LeChat, in simple terms we know that when at equilibrium a system will shift away from any added component and toward anything that is removed. Since in an exothermic rxn heat is a product, heating it At equil will push that left.

As for keq, eq is your hint there temp will definitely affect it, but now you're in keq speak, meaning equilibrium.

 

[TheMightyBoosh]

Understood, thanks for the clarification.