For a reaction at 37 celcius, delta H naught = -10kcal/mol and delta S naught =

Question

For a reaction at 37 celcius, delta H naught = -10kcal/mol and delta S naught = -40 cal/mol-K.

(a) Will the reaction (under standard state conditions) proceed forward or backward?

(b) What is the standard free energy change for the forward reaction?

(c) Which component of the standard free energy, enthalpy or entropy, provides a larger driving force ar 37 celcius for the reaction in (b)?

2. Fructose-1-phosphate can be converted to fructose and phosphate through the following hydrolysis reaction: Fructose-1-phosphate -> fructose + Pi (reversible reaction; I just don't how how to put <-arrow on the reaction)

A reaction is initiated at 30 celcius degree by dissolbing 0.2M fructose-1-phosphate in an aqueous solution. When the reaction reaches equilibrium the concentration of fructose-1-phosphate is 6.53*10^-5M.

(a) Calculate Keq for this reaction at 30 celcius degree.

(b) Calculate delta G naught for this reaction at 30 celcius degree.

(c) TRUE or FALSE? The large favorable change in standard free energy from this reaction indicates that it will reach equilibrium activity.

Explanation / Answer

1)deltaG=deltaH-TdeltaS

=2400cal/mol

==>deltaG>0 therefore reaction proceed backward

b)2400cal/mol

c)standard free energy(deltaG)

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