Week 1- 5 Q’s With Answers

Question 1 _T"\C(VY\OC"\CM. (’\(S{' LOU~), OUpMd H An electric heater is used to supply 2.50 kJ to a 25.0 g sample of H,O originally at 22.0 OC. Calculate the final temperature of the water sample. The specific heat capacity of water is 4.184 J°C~1g-1. 22.0°C N222°C = () 45.9 °C )217.0 °C 1261.0°C W Hide question 1 feedback q=2500J=mC AT m=250¢g C, = 4.184 J/(g °C) Tinitial = 22°C Question 2 0/ 1 point For the reaction 2 NO,(g) = 2 NO(g) + O,(g) AH° = 114.2 kJmol~1 at 298 K. At 1 atm pressure, this reaction will ) not be spontaneous at high temperatures. ' be spontaneous at all temperatures. »() be spontaneous at high temperatures. 'not be spontaneous at any temperature. ' be driven by the enthalpy. \E\ Hide question 2 feedback Change in enthalpy is positive. Change in entropy is positive (i.e., producing more moles of gas) AG = AH -TAS Since the system is entropy driven, the reaction will be spontaneous at a high temperature (i.e., AG < 0) when T is large enough for TAS to become larger than AH. The second term will then be more negative than the first term is positive.

Question 3 0/ 1 point Which of the following has a standard enthalpy of formation equal to zero? - Cle) ) 12(8) ) Na(l) = () Hel) ) S(g) \E\ Hide question 3 feedback Standard enthalpies of formation which equal zero are for elements in their standard state and H*(aq) Question 4 0/ 1 point If the standard enthalpy of combustion of ethanol, CoH5OH(l), is -1368 kJ mol~1, calculate the heat output if 1.00 kg of ethanol is burned. 163.0 x 103 kJ 1.37x10%kJ = ()29.7x10%kJ 337K 1133.7x10%kJ \E\ Hide question 4 feedback 1.00 kg = 21.7 moles g=nxAH g=21.7 mol x 1368 kJ/mol

Question 7 0/ 1 point Calculate the enthalpy change for the combustion of 10.0 g of Mg(s) given the standard enthalpy of formation of MgO(s), - 601.70 kJ mol-1 ' -601.7 kJ ' -123.8 kJ ' -60.17 kJ ' -495.0 kJ = () -247.5k) ¥ Hide question 7 feedback The std. enthalpy of formation of MgO is the same as the enthalpy change for combustion of Mg(s). Determine the number of moles of Mg and multiply by 601.70 kJ/mol Question 8 0/ 1 point Calculate the work needed to make room for products in the combustion of 1 mole of CHyl(g) to carbon dioxide and water vapor at STP (1.000 L atm = 101.3 J). ' -2.26 kJ = () no work is needed ) =6.79 kJ ' -11.3 kJ -4.52 kJ \E\ Hide question 8 feedback Reaction is: CHy(g) + 2 O5(g) = CO5(g) + 2 H,O(g) An=3-3=0 w=-PAV=AnRT =0

Question 9 0/ 1 point Calculate AU for a system that does 150 kJ of work on the surroundings when 300 kJ of heat are released by the system. ) +450 kJ )0 kJ + 150kJ = () -450kJ ' -150 kJ \:’ Hide question 9 feedback AU =q+w When a system does work ON the surroundings, the sign of w is negative (i.e., an expansion). The system must use some of its energy to do work. When heat is released, the system is exothermic. The system releases some of its energy to the surroundings in the form of heat. Question 10 0/ 1 point In a certain exothermic reaction at constant pressure, AH = -150 kJ and 100 kJ of work was required to make room for products. What is AU? » () -250kJ =50 kJ -+ 50kJ 0kJ ' +250 kJ \i’ Hide question 10 feedback AU=q+w At constant pressure, q = AH = -150 kJ Work required to make room for products refers to an expansion process (i.e., w is negative). The system must use some of its energy to push back the surroundings.

Question 13 0/ 1 point For the vaporization of water at 100 °C, ) ASsurroundings 1S greater than ASq ctem ) ASsurroundings IS Positive ) (ASsurroundings * ASsystem) is less than zero w( ) ASsurroundings = _Assystem ) ASsurroundings 1S 1€ss than ASgyctem ‘EW Hide question 13 feedback For vapourization at a substances normal boiling temperature, means system is at equilibrium and process is proceeding reversibly. At equilibrium, AS iverse = O Question 14 0/ 1 point If the enthalpy of fusion of water at its normal melting point is 6.00 kJ mol~1, calculate AS° for melting 1 mole of water at this temperature. ) -22.0 J K 1mol-1 ) -6.00x10%J K Imol? ) +6.00 x 103 J K~1mol-1 1 -20.1 J K 1mol~1 = 220JK 1 moll E\ Hide question 14 feedback fusion = melting normal melting point of H,O = 273K At the melting point, the system is at equilibrium (reversible process). ASap = dreversible/ T = AHyap/T Question 15 0/ 1 point The reaction of 50 mL of hydrochloric acid with 50 mL of sodium hydroxide resulted in a 1.15 °C increase in temperature in a calorimeter with a heat capacity of 4.81 kJ oc-1, What is the heat output of this neutralization reaction? » () 5.53 kJ )1.32kJ 1 0.553 kJ ,4.18 kJ 0.418 kJ i\ Hide question 15 feedback q=CxAT C=4.81kJ/°C AT =1.15°C

'The,(modjmmics| En{'ho.lpa ( En-l-*roPa Question 1 0/ 1 point In a certain reaction at constant pressure, AH = +225 kJ and 35 kJ of work was required to make room for products. What is AU? =260 kJ =190 kJ v +260 kJ ) OklJ = () +190 kJ Question 2 0/ 1 point A calorimeter containing 200 mL of water was calibrated by carrying out a reaction which released 15.6 kJ of heat. If the temperature of the calorimeter rose 3.25 °C, what is the heat capacity of this calorimeter? = () 4.80k°C! )2.58 kJ°c1 1837kj°ct 0960 kJoct 240k °c? Question 3 0/ 1 point A compound has an enthalpy of sublimation of 86.0 kJ mol~1 and an enthalpy of melting of 7.2 kJ mol~1. Calculate the enthalpy of vaporization of this compound. = () +78.8 kJmol~1 ' -93.2 k) mol~1 ' +93.2 kJ mol~1

Consider the reaction 2MnO(s) + Oy(g) @ 2 MnOy(g) AH®=-269.6kJ mol-1 If the standard enthalpy of formation of MnOy(s) is -520.0 kJ mol 1, calculate the standard enthalpy of formation of MnQO(s). = () -385.2 kJ mol'l ) -654.8 kJ mol1 ' -1309.6 kJ mol1 ) =770.4 kJ mol1 ' -250.4 kJ mol1 Question 8 0/ 1 point When SnOy(s) is formed from the combustion of gray tin, the reaction enthalpy is -578.6 kJ mol~1, and when white tin is burned to form SnO,(s), the reaction enthalpy is -580.7 kJ mol~1. Calculate the reaction enthalpy for Sn(gray) — Sn(white) ) +1159.3 k) mol~1 ' -1159.3 kJ mol~1 -2.1kJmol™! o () +2.1kJ mol™1 Question 9 0/ 1 point List the following in order of decreasing molar entropy at 298 K. HClI(g), Cly(g), HCl(aq), Cl(g) ) Cl(g) > Cla(g) > HCl(g) > HCl(aq) ) HCl(g) > HCl(aq) > Cla(g) > Cl(g) = () Cl2(g) > HCI(g) > Ci(g) > HCl(aq) ) Cla(g) > Cl(g) > HCI(g) > HCl(aq) N HCl(aq) > Cl(g) > HCl(g@laClale) Question 10 0/ 1 poir Which of the following reactions has the largest positive molar entropy change? () H20(s) = H20(g) ) CHylg) + 2 O5(g) — COx(g) + 2 Hy0(g) ) PCls(g) — PCl3(g) + Cla(g) ) Na(g) + 3 Ha(g) = 2 NHg(e) = () KClIO4(s) + 4 C(s) = KCl(s) + 4 CO(g)

Gilbees GDO/CO‘O Question 1 0/ 1 point Determine AG°for the following reaction: CHylg) + 2 Og(g) = COx(g) + 2 HyO() Substance A_Gfi(kJ/moI)_ CHyl(g) -50.72 O,(8) 0 CO5(g) -394.4 H,O()) -237.4 -581.1 kJ = () -818.5kJ ) 131.1 kJ ' -919.9 kJ - -682.5 kJ Question 2 0/ 1 point Consider the following reaction: 3 C(s) + 3 Hylg) = C3Hylg); AHP =20.41kJ moll; A8 =-131.6 J K1 mol'lat 298 K What is the equilibrium constant at 298 K for this reaction? )2.8x10%10 )2.6x 1074 ) 1.0 =»()3.5x10711 1.3x 1077

Question 4 0/ 1 point For the reaction CaCOg3(s) = CaO(s) + O,(g) at 1 atm pressure, the values of AH and AS are both positive, and the process is spontaneous at high temperatures. Which of the following statements about this reaction is true? = () The change in entropy is the driving force for the reaction. ) The reverse reaction is endothermic. . The process is exothermic at high temperatures and endothermic at room temperature. ) AG at room temperature is negative. ) The reverse reaction is nonspontaneous at room temperature.

Which of the following statements is true concerning the reaction below? CHy(g) + No(g) = HCN(g) + NH3(g); AH° =164.1kJmoll; AG°=159.1kJ mol'lat 298 K ) It is nonspontaneous at all temperatures. ) It is spontaneous at all temperatures. ) Itis spontaneous at relatively low temperatures only. »() It is spontaneous at relatively high temperatures only. ) Itis at equilibrium at 298 K. ¥ Hide question 5 feedback Given: CHy(g) + No(g) = HCN(g) + NH3(g); AH° = 164.1 kJ mol?! AG® = 159.1 kJ mol'! at 298 K Want: True statement Concept(s): AG°® = AH® - TAS® Reactions support the formation of products when the AG® value is negative. Reactions are "enthalpy driven" when the sign of AH is negative, as no external heat is required for the reaction to proceed in the forward direction. Reactions are "entropy driven" when the sign of AS is positive, as that aligns with the 2nd law of thermodynamics, i.e., AS hiverse > O for a spontaneous process. If AH is negative and AS is positive, the rxn will favour the production of products at all temperatures, as both terms are supporting that rxn direction. If AH is positive and AS is negative, the forward reaction will not be supported at any temperature. If the two signs match, then one term is supporting the products and the other is not. Temperature is then the deciding factor to influence the outcome. Analysis: Determine the value for AS° based on the information provided and then identify which statement is true.

Question 8 Consider this reaction. H20(g) + 0.5 02(g) —> H202(g) AMH° = +105.5 kJ mol 1 What can you determine about the reaction's spontaneity? ) Spontaneous at all temperatures. Spontaneous at low enough temperatures. ) Spontaneous at high enough temperatures. = () Not spontaneous at any temperature. Question 9 Consider this reaction. CgHglg) + 7.5 O5(g) = 6 CO4(g) + 3 H,O(g) AH° = -3135 kJ mol ! What can you determine about the reaction's spontaneity? = () Spontaneous at all temperatures. ) Spontaneous at low enough temperatures. ) Spontaneous at high enough temperatures. Not spontaneous at any temperature.

Question 10 Consider this reaction. NO(g) + 0.5 Cl,(g) — NOCI(g) AH° = -38.54 kJ mol1 What can you determine about the reaction's spontaneity? . Spontaneous at all temperatures. = () Spontaneous at low enough temperatures. ) Spontaneous at high enough temperatures. ) Not spontaneous at any temperature. Question 11 Consider this reaction. 2 NH3(g) = N(g) + 3 Ha(g) AHC = +92.22 kJ mol? What can you determine about the reaction's spontaneity? ) Spontaneous at all temperatures. » Spontaneous at low enough temperatures. = () Spontaneous at high enough temperatures. ) Not spontaneous at all temperatures.

Question 15 0/ 1 point Consider this reaction. 2 Al(s) + 6 H*(aq) — 2 AlI%*(aq) + 3 Hy(g) The standard Gibbs energy of formation for the aqueous aluminum ion is -485 kJ/mol. What is the standard Gibbs energy of this reaction? +485 kJ/mol = () -970 kJ/mol 485 kJ/mol ) +970 kJ/mol Question 16 0/ 1 point Note the following reactions. The standard Gibbs energy of reaction values are for 25 °C: SO,(g) + 3 CO(g) = COS(g) + 2 CO,(g) A,G° = -246.4 kJ/mol CS,(g) + H,O(g) — COS(g) + H,S(g) AG° =-41.5 kJ/mol CO(g) + H,S(g) — COS(g) + Ho(g) AG° =+1.4 kJ/mol CO(g) + H,O(g) — CO5(g) + Ho(g) A,G° = -28.6 kJ/mol With these data, determine A,G° for the following reaction. COS(g) + 2 H,O(g) — SO5(g) + CO(g) + 2 Ho(g) + 303.6 kJ/mol -189.2 kJ/mol ' -303.6 kJ/mol o +189.2 kJ/mol

What is the equilibrium constant at 298 K for this reaction? N2O(g) + 0.502(g) = 2 NO(g) The standard Gibbs energy of formation of N,O(g) is +104.2 kJ/mol while that for NO(g) is +86.55 kJ/mol. = ()8.38x1013 10.97 11.03 11.19 x 1012 11.24 x 108 18.06 x 1074 Question 18 0/ 1 point What is the value of the equilibrium constant for this reaction? 2N204(g) + O2(g) = 2N205(g) The standard Gibbs energy of formation for NoOy4(g) is +97.89 kJ/mol while that for N,Os(g) is +115.1 kJ/mol. ) 1.08 x10° 19.63x10°4 w( )9.27 x1077