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### k & K

Posted: Fri Mar 08, 2019 8:30 am
What is the difference between k in kinetics and K in equilibrium/ thermo?

### Re: k & K

Posted: Fri Mar 08, 2019 8:51 am
k in electro should be Boltzmann's constant while k in kinetics is a number unique to each reaction scaling how quickly the reaction proceeds.

### Re: k & K

Posted: Fri Mar 08, 2019 10:00 am
To add to the previous answer, K in thermodynamics is the equilibrium constant, which describes the concentrations of reactants and products at equilibrium, and doesn’t have anything to do with the speed of the reaction.

### Re: k & K

Posted: Fri Mar 08, 2019 11:14 am
k is a rate constant dependent on the reaction you have. K is the equilibrium constant.

### Re: k & K

Posted: Fri Mar 08, 2019 12:34 pm
how do you know when to use it as a Boltzmann's constant or a rate constant?

### Re: k & K

Posted: Fri Mar 08, 2019 12:34 pm
or are those the same?

### Re: k & K

Posted: Sat Mar 09, 2019 5:26 pm
isn't boltzmann's constant used to calculate entropy with the number of possible states of a molecule and the number of molecules?
rate constants and boltzmann are entirely different things.

### Re: k & K

Posted: Sat Mar 09, 2019 11:27 pm
k is Boltzmann constant, which is equal to 1.38064852 × 10^-23 m2 kg s^-2 K^-1. K is the equilibrium constant of a reaction, which is calculated by the concentration of the products over concentration of the reactants.

### Re: k & K

Posted: Sat Mar 09, 2019 11:38 pm
k is the Boltzmann rate constant while K is the equilibrium constant for a certain reaction.

### Re: k & K

Posted: Sat Mar 09, 2019 11:48 pm
I am sure the question will tell us to use the k as a rate constant in kinetics in the question, whereas the k Boltzmann constant will be used to calculate the entropy. K will be given in an acids and base question. It is based off of context.

### Re: k & K

Posted: Sun Mar 10, 2019 12:44 pm
The difference is that k is Boltzmann's constant, while K is the equilibrium constant for a given reaction.

### Re: k & K

Posted: Sun Mar 10, 2019 1:16 pm
In kinetics, k is the rate constant, whereas K is the equilibrium constant. Also, K = k/k', which means the the equilibrium constant is equal to the ratio of rate constants of the forward (k) and the reverse (k') elementary reactions.

### Re: k & K

Posted: Wed Mar 13, 2019 10:16 am
lower case k is the boltzmann constant and upper case K is the equilibrium constant

### Re: k & K

Posted: Wed Mar 13, 2019 2:35 pm
k is the rate constant so it's basically an indicator of how fast the reaction will happen. K is the equilibrium constant that tells you the ratio of products to reactant concentrations when the reaction is at equilibrium. Based on the relationship between k and K, we can determine that the forward reaction's rate constant, k will be larger than the reverse reaction rate constant if K (equilibrium constant) is larger than 1.

### Re: k & K

Posted: Wed Mar 13, 2019 8:42 pm
It is also important to note that K can be figured out by taking the ratio of the forward reaction K to the reverse reaction K

### Re: k & K

Posted: Fri Mar 15, 2019 10:33 pm
Will there be a clear indication of 'k' and 'K' on the final? Such as bolding one over the other. Thanks.

### Re: k & K

Posted: Sat Mar 16, 2019 12:32 pm
Melissa Villanueva1K wrote:Will there be a clear indication of 'k' and 'K' on the final? Such as bolding one over the other. Thanks.

Do you mean on the constants and equations sheet?

### Re: k & K

Posted: Sat Mar 16, 2019 2:02 pm
Melissa Villanueva1K wrote:Will there be a clear indication of 'k' and 'K' on the final? Such as bolding one over the other. Thanks.

I think in order to avoid confusion, it may be written out for us along with the variable e.g. "the rate constant, k,..."

### Re: k & K

Posted: Sat Mar 16, 2019 2:23 pm
K is the equilibrium constant and kr is the rate constant, symbolizing different constant values with different meanings, applied in different equations but they are related. K is equal to the overall forward rate constant over the reverse rate constant because at equilibrium the forward and reverse rates (not the constants) are equal.