Discussion Week 7 + Tut7

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Chemistry

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Nov 24, 2024

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Week 7 Discussion + Tutorial S22 CHEM 110A May. 9** May. 15* 1 Things learnt after MT1 Entropy
= 2A0F (')’1-' LT 1= RT= HREK 2 Microscopic picture 2.1 PM1-Q1 aH = -[ootm™ Tertiary amines can form a bond with certain metals and metalfxide sur- faces. The formation of this bond is exothermic and releases 100 e~ of heat ~(00 &h"' per bond. You design a molecule with four tertiary amine limbs to adsorb to L '.-b.,l surfaces. This molecule can adsorb either with four (1 way), three (4 way), two - e e (6 way), or one (4 way) “limbs". The head does not stick to the surflace, (a) Calculate the partition function Z for the four energy states shown in the figure. Use kgT = 207 emn™' You are provided with the possible ways to achieve each state in the figure, —_ (b) What percentage of molecules will be attached with one imb? What ( percentage will be attached with four limbs? <) -1 * (c) Calculate the equilibrium constant for a molecule going from one limb - (’000" attached to four limbs attached. s (d) Without doing any calculations, explain what will happen to this equi- 4.- é’& librium constant if temperature is raised? E: (a) 2= i;AASe A ~ -6, o, TR o~ BOAR (e o ot + G et/ | s (b) / twb D _ 4 oo categ S (= 1"%op . Ne” 4-€ = 0% E’ - “). ,6‘ 0 *(e) IS4 > K= '}}72 %21-0-7 d) AG= A"'/'%—é-@w 48 s Jowored ! v::a: # mrywstodes
2.2 PM2-Q1 The molecules have [thioll-like functional groups that allow them to cova- lently bind to gold on a 2-D surface. Consider a 2-by-2 portion of this surface, « 0 hydrogen bonding = 2 configurations (highest E) « 2 hydrogen bonding = 12 configurations (intermediate E) -’@4 hydrogen bonding = 2 configurations (lowest E) A (a) Shown above are all possible arrangements in this 2 x 2 grid. If there is no enthalpy associated with the hydrogen bonds, what is the probability of finding 2 x 2 system evenly distributed (the “four hydrogen bonds®™ section)? (b) In reality, formation of one hydrogen bond is exothermic and releases 0’ 300 em~". Assuming “no hydrogen bonds” corresponds to 0 joules, calculate - the partition function at T = 208 K for hydrogen bonding in the 2 x 2 grid. %\ vo {(c) At T = 298 K, what is the probability of achieving no hydrogen bonds? s’ w‘ What is the probability of achieving four hydrogen bonds” (d) No calculation necessary. To increase the probability of the “evenly distributed” state with four hydrogen bonds, should you increase or decrease temperature’
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2.3 PM1-Q6 and Q7 Consider the following endothermic reaction: (‘a('();u'.' = (‘flol.) + ("02!_91' Decreasing the amount of CaO in the reaction vessel will (a) Shift the reaction towards the products (L) Shift the reaction towards the reactants @lnw no effect on the reaction é—— 9' &&3 Mc ‘) Decreasing the temperature of the reaction vessel will (a) Shift the reaction towards the products Shift the reaction towards the reactamnts (c) Have no effect on the reaction N a\ow-b AH >0 R,: P AHP"AHR, >0 T it ~ o P yeotk /fie@ e/@ . ¥
3 Macroscopic picture 3.1 PM1-Q3 5kJ of hea t;w through a lm! n ¢ where the hot source is at 300 °C and the »Il l 25 (' The l gine n;wmlm ul muimmn efficiency. How mm'! ropy is generated a l ll ink, assuming the temperature of he cold xink 'us constant? ~ T ] 30 AQ = éé:“_c 000 c T ‘g" =+ L 3 205 -w-3.= .w-%j 8% = —2.68] =- g = 5c>;+t‘i‘3)k (=-4%5) 3-3%
3.2 PM1-Q4 The bond between carbon and iodine in organic molecules is quite weak in comparison to carbon-hydrogen or carbon-carbon bonds, The reported bond dissociation enthalpy for the cleavage of the C-1 bond in iodomethane is AH® = 240k /mol, CH3yl CH, + I, This enthalpy change is the standard enthalpy and therefore only applicable at room temperature. What is the enthalpy change at 1000 K under constant pressure conditions? Use: 5 cecuyt = 4R, cpen, =4R, cpy = ER' Ctote et &F H. AH((wok) = AH + 4ArH = AHO ¢+ (A”r“’-‘”fl) = AHO + (M (g 8T* N CaT = Nerg Corgz 4T) AHO + (Con + G = Congr ) AT - 349% + [ gRt Le- u).((ooo—lfif)k = Jm;‘% + l@é% Q58 RS fnel ] 1\
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3.3 PM1-Q8 Standard entropies at room temperature can be Yositive only 5) Negative only 0 (c) Positive or zero S.—-— (d) Negative or zero a'b T:OK (e) Positive, negative, or zero