Question

October 11, 2019 CHEM 1252 Problem Set 11 Textbook Reading Assignment: 12.5-8 Additional Practice Problems: 12.43-44,46-48,50-55,57,59-61,63,65,67,73,75,81 1. A certain reaction has a All=-10.5 kJ and a AS +50.5 3/K (a) Explain whether or not the reaction is endothermic or exothermic. (b) Explain if the reaction is leading to an increase, decrease, or no change in disorder of the system. (c) Is the reaction spontaneous at 0 °C? 2. Use data from Appendix 4 (Table A4.3) to calculate Al, AS", and AG at 298 K for each of the following reactions. In each case show that AG Al"-TAS", and tell whether the reaction is spontaneous at 298 K under standard conditions. Be mindful of the units! (a) 2Ca(s) O2(g) 2Ca0(s) + дHP: AG (b) CH CO H() CHa(g) + CO2(e) AH AGP 3. The reaction in 2b can be classified as decarboxylation because carbon dioxide is produced as a product. Applying the calculated data from 2b, what is the minimum temperature at which this reaction will be spontaneous under standard conditions?

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Answer 1

dH = -10.5 kj
a) The negative sign for enthalpy change indicate the reaction is exothermic. That
is heat is released during the process

b) dS= +50.5 J/K
Positive sign for dS indicate there is an increase in entropy. Entropy
is a measure of disorder . More the entropy higer the disorder
So there is an increase in entropy

C)
we have free energy change dG = dH -TdS
dH = -10500 J, dS= +50.5 j/k
when dG <0 the reaction is spontaneous
Temperature T= 0 +273 K = 273 K
dG = -10500 J - [273 K* 50.5 j/k]
= - 24286.5 j
negative sign indicate the reaction is spontaneous at 0 degrees
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2Ca(s) + O2 (g) ---> 2CaO(s)

dH = dHf(products) -dHf(reactant)
= 2*dHf(CaO) - 2dHf(ca) -dHf(O2)
= 2*[-635.09 Kj] -0
= -1270.18 KJ = -1270180 J

dS = ds(products) -ds(reactant)
= 2*dS(CaO) - 2*dS(ca) -dS(O2)
= 2*[39.75]- 2[41.42 ] -[205.138]
= -208.478 J/K
T= 298 K

dG = dH-TdS =-1270180 J-[298K*-208.478 J/K] =-1208.07 KJ
negative sign indicate reaction is spontaneous
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