Also, if you would be so kind as to describe each reaction (part A,B,C) as spont

Question

Also, if you would be so kind as to describe each reaction (part A,B,C) as spontaneous or nonspontaneous.

MasteringChemistry: Thermochem 2 - Google Chrome https://session.masteringchemistry.com/myct/itemView?assignmentProblemID-48069034&offset-next; Chem 116 Fall 2015 Thermochem 2 Item 6 Help Close Resources « previous | 6 of 7 | next » Item 6 Part A 320 K Consider the following reaction CaCO3 (s) CaO(s) + CO2 (g) Estimate Go for this reaction at each of the following temperatures. (Assume that AH° and AS° do not change too much within the given temperature range.) kJ Submit My Answers Give Up Part B 1100 K kJ Submit My Answers Give Ujp Part C 1455 K A2 kJ 11:20 AM 11/4/2015

Explanation / Answer

Solution :-

Balanced reaction equation

CaCO3(s)   ------ > CO2(g) + CaO(s)

Using the standard enthalpy of formation and standard entropy values lets first calculate the enthalpy and entropy change of the reaction

Calculating the enthalpy change

Delta H reaction = sum of delta Hf product - sum of delta Hf reactant

                              = [(CO2*1)+(CaO*1)] – [CaCO3*1]

                              = [(-393.5*1)+(-635.1*1)]-[-1206.9*1]

                              = 178.3 kJ per mol

Calculating the entropy change

Delta S reaction = sum of delta S product - sum of delta S reactant

                              = [(CO2*1)+(CaO*1)] – [CaCO3*1]

                              = [(213.7*1)+(38.2*1)] –[ 92.9*1]

                              = 159 J per mol

Lets convert it to the kJ

159 J per mol * 1 kJ /1000 J =0.159 kJ/mol

Now lets calculate the Delta G change at the each given temperature ‘

1)At 320 K

Delta G = Delta H – (T*delta S)

              = 178.3 kJ – (320 K * 0.159 kJ/mol K)

              = 127.42 kJ/mol

Since delta G is positive therefore reaction is non spontaneous at this temperature

2)Calculating at 1100 K

Delta G = Delta H – (T*delta S)

              = 178.3 kJ – (1100 K * 0.159 kJ/mol K)

              = 3.40 kJ/mol

Since delta G is positive therefore reaction is non spontaneous at this temperature

3) Calculating at 1455 K

Delta G = Delta H – (T*delta S)

              = 178.3 kJ – (1455 K * 0.159 kJ/mol K)

              = -53.05 kJ/mol

Since delta G is negative therefore reaction is spontaneous at this temperature

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