I don\'t understand the difference in questions B & C - how would you calculate

Question

I don't understand the difference in questions B & C - how would you calculate each?

And help with this one, too, please:

THANK YOU VERY MUCH!

The compound BrCl(g) is produced by mixing bromine gas and chlorine gas at 298 K. The mixture reaches equilibrium in a very short time. Br2(g) + Cl2(g) 2BrCl(g) What is the numerical value of the standard entropy change, delta S degree, at 298 K for this reaction? Include units. What is the numerical value of the standard entropy change of formation, delta S degree, for BrCl(g) at 298 K? Include units. Calculate the standard entropy change of formation, Delta S f degree, for glucose, C6H12O6(s) at 298 K. Specify units. Calculate the standard free energy change of formation, Delta G f degree, for glucose, C6H12O6(s) at 298 K. Specify units. At 298 K, glucose burns in oxygen to form only CO2(g) and H2O(l). Calculate the expected value of the standard enthalpy of combustion, Delta H comb degree, for glucose, C6H12O6(s). Specify units.

Explanation / Answer

1.

B)

Detla S = S products - S reactants

Delta S = [2 x BrCl] - [S of Br2 + S of Cl2]

=[2 x 239.7] - [152.2 + 222.8]

= 101.4 J/mol K

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For formation the reaction is for 1 mol

1/2Br2(g) + 1/2Cl2(g) <----------> BrCl(g)

Thus, delta Sf = Delta S / 2 = 101.4 / 2 = 50.7 J/mol K

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2.

B)

6 CO2(g) + 6 H2O(l) --------------> C6H12O6(s) + 6 O2(g)

Delta S = [S of C6H12O6 + 6 x S of O2] - [6 x S of CO2 + 6 x S of H2O]

= [212.1 + 6(205.0)] - [6 x 213.6 + 6 x 69.96]

= -259.26 J/mol K

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C)

Delta H = [H of C6H12O6 + 6 x H of O2] - [6 x H of CO2 + 6 x H of H2O]

= [-1273.02] - [6 x -393.5 + 6 x -285.83]

= 2802.96 kJ/mol

Delta G = Delta H - T. Delta S

Delta G = (2802.96 x 1000) - (298) (-259.26 )

= 2880219.48 J /mol

= 2880.22 kJ/mol

D)

C6H12O6 + 6O2 <--------------> 6CO2 + 6H2O

Delta H = [6xHCO2 + 6 x H H2O] - [H of C6H12O6 + 6 x H of O2]

= [6 x -393.5 + 6 x -285.83] - [-1273.02]

= -2802.96 kJ/mol

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