A combustion reaction releases 783 kJ of energy at 25 degree C (a) Calculate the

Question

A combustion reaction releases 783 kJ of energy at 25 degree C (a) Calculate the change in entropy of the surroundings, Delta S_ (b) Calculate the change in entropy if the reaction is done at 400 degree C. a) A reaction at a high temperature has a negative value for enthalpy and a negative value for entropy. Is this reaction spontaneous (b) What about at a low temperature? Consider the following information A + B rightarrow C Delta H = - 40.5 kJ; Delta S = 342.1 J/K 40 degree C (a) Just by looking at the signs of the values for entropy and enthalpy, is this reaction spontaneous or nonspontaneous? (b) Calculate the change in free energy of the reaction, Delta G. (c) Is there a temperature where the reaction will reverse in spontaneity? (a) Use the standard entropy values (Delta S_ r degree) in the Appendix to solve for Delta S degree; of each reaction at 25 degree C (b) Solve for Delta G degree _ for both reactions as well using the standard free energy values. (Delta G degree _) 2H_2 O rightarrow 2H_2(g) + O_2 (g) Al_2 O_3(s) + 3 H_2 (g) rightarrow 3 Al (s) + 3 H_2 O(g) N_2(g) + O_2 (g) rightarrow 2 NO (g) Delta H degree = 180.7 kJ; Delta S degree = 24.7 J/K 25 degree C (a) Calculate the standard free-energy change for the reaction at 25 degree C. (b) Calculate Delta G at 500 degree C using the Delta G degree value calculated from part a.

Explanation / Answer

(1)

deltaSsys = - qsys/T = - 783 / 298 = - 2.63 kJ/K.mol

DeltaSsurr = - qsys/T = - (- 783 / 298 ) = + 2.63 kJ/K.mol

DeltaSsurr = - q sys / T = + 783 / 673 = 1.16 kJ/K.mol

(2)

We know the relation of Gibss ffree energy with the enthalpy and entropy as,

deltaG = deltaH - TdeltaS

For a reaction to be spontaneous, deltaG should be negative, i.e. deltaG < 0

If deltaS is negative, - TdeltaS is positive.

Hence at high temperatures, though deltaH is negative, combined value of detaH-TdeltaS will be positive. Hence the reaction is non spontaneous.

But low temperatures, though deltaS is negative, i.e. TdeltaS term is positive, the collective effect of deltaH - TdeltaS will be negative. Hence at low temperatures the reaction is spontaneous.

(3) for a reaction to be spontaneous, deltaG = deltaH - TdeltaS = - ve.

(a) Since deltaH is negative and deltaS is positive, deltaG is negative always. SO, the reaction is spontaneous.

(b)

deltaG = - 40.5 - 313(0.3421) = - 147.6 kJ/mol

(c) No at all temepratures the reaction is spontaneous.

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