What Is the Entropy of a Solution Containing Methane and Ethane?

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The discussion centers on calculating the absolute third law entropies of a solution containing one mole each of methane and ethane at 25 degrees Celsius and 1 atm pressure. The provided absolute third law entropies for methane and ethane are 186.19 J/(mol*K) and 229.49 J/(mol*K), respectively. Participants are encouraged to attempt the solution or ask specific questions for assistance. It is suggested to review the section on 'entropy of mixing' in the relevant textbook for better understanding. The focus remains on applying ideal gas behavior to determine the overall entropy of the mixture.
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Homework Statement



At 25 degrees celcius and 1 atm pressure, the absolute third law entropies of methane and ethane are 186.19 J/(mol*K) and 229.49 J/(mol *K) respectively in the gas phase. Calculate the absolute third law entropies of a solution containing 1 mole of each gas. Assume Ideal gas behaviour.
 
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On these forums you need to attempt the solution or ask specific questions for us to be able to help you. What exactly are you having trouble with / don't you understand about this question?
Perhaps read the section in your text on the 'entropy of mixing'.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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