When sucrose is hydrolyzed to glucose and fructose, what are the changes in enthalpy, entropy, and free energy?

in Hydrolyze Reviews




The answer isenthalpy decreases entropy increases free energy decreasescan anyone explain how.


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billiejean254 January 11, 2011 at 8:51 pm

free energy = enthalpy – entropy (temp)

Enthalpy decreases – just means that heat is given off in the reaction, it varies from reaction to reaction, there’s really no way for you to know this without testing it or looking it up in a book. (or without knowing a lot of info about all the products and reactants)

entropy increases – easy one. entropy is the randomness of the universe. so if one molecule turns into two, it becomes more random, so entropy goes up. If two molecules become one entropy goes down.

lastly use the equation given above if enthalpy is negative and entropy is positive, no matter what values you assign them the free energy will be negative (or decrease) also the temp doesn’t matter either since it is in Kelvin and can never be negative.

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dudette January 15, 2011 at 4:13 am

if enthalpy change < 0, it is an exothermic reaction which involves bond formation, i.e. energy is given off. hydrolyzing sucrose to glucose and fructose involves the breakage of the glycosidic bond between glucose and fructose. however, it involves the formation of glucose and fructose as well by the attachment of H and OH groups to the respective sugars, which could probably cancel out the energy absorbed for bond breakage. so the enthalpy of the system decreases, or deltaH is negative.

entropy or S refers to the randomness of molecules in a system. entropy increases as there are now two sugar molecules as compared to one. therefore the number of ways in which these molecules can be distributed in a random fashion increases - i.e. the randomness of the molecules in the system increases. so the entropy increases, or deltaS is positive.

gibbs free energy is given by the equation deltaG = deltaH - TdeltaS. we know that deltaH is negative and deltaS is positive. so we substitute in the signs...

delta(-H) - T(deltaS)
= -deltaH - TdeltaS

negative value minus a negative value gives you a negative value. therefore, deltaG is negative, or free energy decreases.

hope this helps.

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