Question

CH (9) H20 (9) 186 1189 0,09 205 ThermodyWUM 301) Use the data in the foblo 600W- 1 0 Hon Compound AS CT mol KP write the b0/0 for the combustion of methane CH49 (1 mark) 6) ootmine the standard entropy Change As for the combustion of 1 mole of 1C0₂ (9) methane, CH (9) (note ollroactants and products in the gaseous Store) 13 marks) ii) Exploin the following process: the entropy chong for freezing ethanol is less thon zoro a Thi entropu chonge for iroporating home ater than aero room tom poroture greater th on 2 Pro following process is sponFonous Or note (3 morts) ml.I dontify whether the following process is spontaneo 01 Irlaub mits D/2N09) +0,69) 2 N0279) (1 mark) Iredict whether the intropy change of the sustom in och of the following actions is positive or no go live. - 2H21g) + O₂ → 2H2O1.. CN Hucis NH3(9) + HCl (1 marks) inocombustion of ethanol 07 298 Kis given by the reaction 6010W GH₃0 H 19 + 30₂ (9) 2102(g) + 3 H₂O CIL The followings ore stondord molar entropy Of Subs+o0 cos at 29€* C2HOH 386/I/mol. 02 205 Ilmolk 2/3-71 /mol K 61.9J/mol Colculoti thi entropu chonges Q298k using dard above State whether the procoSS IS Sponto nou for the combustion of ethanol Solve the VOIUO 0+ Ago for the following refron. CO2 H20 Or non spontaneous 13marks)

Can you help me answer questions 3)a)i)ii)iii)iv).thank you so much for your help.These questions are based on thermodynamics.I really appreciate your kindness and help

Here are the updated questions.I provide you with bigger and clearer pic for all the subparts.I hope you can help me.thanks a lot.

Answer 1

i.a Balanced reaction of combustion of methane

CH4 + 2O2 = CO2 + 2H2O

ib Determination of standard entropy change

Standard entropy change is calculated based on following formula.

$\Delta S^{0}$ = Sum ( Entropy of products) - Sum (Entropy of reactant)

Sum ( Entropy of products) = Sum (No of moles of product * Entropy of single mole)

= (moles of CO2 * Entropy of CO2 + moles of H2O * Entropy of H2O)

= 1 * 214 + 2 * 189

= 592 J/mol K

Sum ( Entropy of reactant) = Sum (No of moles of reactant * Entropy of single mole)

= (moles of CH4 * Entropy of CH4 + moles of O2 * Entropy of O2)

= 1 * 186 + 2 * 205

= 596 J/mol K

$\Delta S^{0}$ = 592 - 596

$\Delta S^{0}$ = -4 J/molK (ANS)

ii.

To determine entropy changes in the case, we examine wether the number of microstates in the system increases or decreases.If there is increase in the number of microstates, entropy change will be positive.

If there is decreases in the number of microstates, entropy will be negative.

a. For Entropy change of freezing ethanol,the molecules of ethanol are held rigid in position,this phase transition reduces the number of microstates and so entroly decreases that is  $\Delta S^$ < 0.

b. Evaporation of bromine increases the number of microstates because the bromine molecule can occupy many more positions in nearly empty space, so $\Delta S^$ > 0

iii

a. Icecube melting is a spontanious process.

b. Oxidation of NO to NO2 is spontanious process.

c. Entropy change of the water formation reaction is negative because the water molecules occupy less spaces compared to that of hydrogen and oxygen molecules in gas phase that is reslulting in the decrease of number of microstates of liquid water.

d. Entropy changes of NH4CL to NH3 + HCl is positive because number of microstates are increasing by formation of two individual molecules of NH3 and HCL.

iv

a. Standard entropy change of the reaction C2H5OH + 3O2 = 2CO2 +3H2O

$\Delta S^{0}$ = Sum ( Entropy of products) - Sum (Entropy of reactant)

Sum ( Entropy of products) = Sum (No of moles of product * Entropy of single mole)

= (moles of CO2 * Entropy of CO2 + moles of H2O * Entropy of H2O)

= 2 * 213.7 + 3 * 68.95

= 634.25 J/mol K

Sum ( Entropy of reactant) = Sum (No of moles of reactant * Entropy of single mole)

= (moles of C2H5OH * Entropy of C2H5OH + moles of O2 * Entropy of O2)

= 1 * 386.1 + 3 * 205

= 1001.1 J/mol K

$\Delta S^{0}$ = 634.25 - 1001.1

$\Delta S^{0}$ = -366.85 J/molK (ANS)

As Entropy change is negative, reaction is non spontanious. (ANS)