Reaction rate and the equilibrium rate

[Levi0sa]

the equilibrium rate involves the constant, K, such as the eq constant = [products]/[reactants]
How does that k relate to Gibbs free energy? Do you simply plug it into the equation -RTlnK? I'm just very confused because my Ochem book says -RTlnK= -2.303RTlogK and I'm not sure where they got this constant -2.303...

Also, does the above k relate at all to the rate constant k where the rate of a reaction = k[product1][product2] .... Or is k just a standard variable to denote constants..

Thanks for help!

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[OpticalBlackOut]

Sorry this is going to sound harsh, but when did we turn into your chemistry tutor?
This is a forum to inquire about health professional career choices, or asking questions about the application process.

Edit: to turn this into a positive to those who are looking into optometry, I havent done a problem like this since second year of undergrad...which has been a while

 

[Levi0sa]

Well .......I guess you never turned Into my chemistry tutor since you didn't really answer my inquiry.

I am interested In optometry and am currently interning at an optometrist facility ..... recently joined SDN because I have seen great answers to undergrad pre-health questions on here before.

 

[Levi0sa]

Sowwy. Will check out other forums.

 

[jroberts]

The constant is to correct for the fact that log and ln are not the same thing. Log is base-10, ln is base-e.

 

[drumstix]

the equilibrium rate involves the constant, K, such as the eq constant = [products]/[reactants]
How does that k relate to Gibbs free energy? Do you simply plug it into the equation -RTlnK? I'm just very confused because my Ochem book says -RTlnK= -2.303RTlogK and I'm not sure where they got this constant -2.303...

Also, does the above k relate at all to the rate constant k where the rate of a reaction = k[product1][product2] .... Or is k just a standard variable to denote constants..

Thanks for help!

[stock SDN advice]
Go with the school that'll cost you less in the long run. It's all the same degrees in the end. Why pay an extra $30,000 when both produce the same eye doctor OD degree?
[/stock SDN advice]