Question

Be sure to answer all parts. Take all the thermodynamic data from your textbook. Assume the data do not vary with temperature. H ,026) at 167.0°C. () Calculate AS only for the reaction H,02 Enter your answer in the box provided. J/K mol (b) is the reaction spontaneous, nonspontaneous, or an equilibrium procesi Select the best single answer O spontaneous o nonspontaneous O an equilibrium process (e) The reaction NH36) + HCIG) - NHC(s) is spontaneous in the forward direction at room temperature. Determine the temperature at which it is no longer spontaneous in the forward directie Enter your answer in the box provided. (a) Determine the boiling point of Brz: Enter your answer in the box provided. < Prex 2 of 10 Next > MacBook Air

Answer 1

(a Thereactionisas follows: H,02(1)–>H,02(8) Calculatethechangein entropy of the reaction as follows: AS men = 5° [H,02(8)]-5°[H,02(1)] = 232.9 J/K.mol-109.6J/K.mol =123.3J/K.mol Similarly, calculate the changein enthalpy of thereaction asfollows: AH°m = AH°[H,02(8)]-AH" [H,03(1)] =-136.1 kJ/mol -(-187.6 kJ/mol ) =+51.5 kJ/mol -- - - -- -- - --- - -- - -- - -- - --- - - -- -- - --- - - - - Calculate the entropy change of thesurroundingsasfollows: AS = -AH°men 7 1kJ -51.5 kJ/mol 1000J 440 K = -117.0 J/K.mol Finally, AS is calculatedas follows: ASuray = AS°un + AS, = 123.3J/K.mol - 117.0 J/K.mol = 6.3 J/K.mol - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (b) The sign of ASuriy is positive. Thus, the processisspontaneous.

(c)The reactionisasfollows: NH (8)+ HCl(8)–>NH,Cl(s) Calculatethechangein entropy of the reaction as follows: AS man = Sº [NH,A(s)]-S [NH (8)]-5 [HCI(:)] =94.56J/K.mol-193.0J/K.mol – 187.0J/K.mol =-285.44J/K.mol Similarly, calculate the changein enthalpy of thereaction as follows: AH°m = AH°(NH,A1(s)]-AH°[NH (8)]- AH°(HCI(8)] =-315.39 kJ/mol -(-46.3kJ/mol)-(-92.3kJ/mol) =-176.79kJ/mol --------------- --- --------- ---- -- Calculatethe entropy change of the surroundings as follows: Asvx = -AĦºwe T Xa _ +176.79J/mol 1000J I 1kJ - +176.79x10°J/mol Finally, AS uy is calculatedas follows: AS war = ASX + ASJUX 0 = -285.44J/K.mol-176.79x109 J/mol _176.79x10²J/mol 285.44J/K.mol =619.4K - 273 =346.4°C

Therefore, thetemperature at which thereactionis nolonger spontaneous in forwardreaction is 346.4°C. - - - - - - - - - - - - - - - (d) Thereaction is as follows: Br(1)–>Br2(8) Calculate the changein entropy of the reaction as follows: 45 mm = 5° (Br2(x)]-5° (Brz(1)] = 245.13J/K.mol-152.3J/K.mol =92.83J/K.mol Similarly, calculate the changein enthalpy of thereaction as follows: AH m2 = AH°(Br2(8)]-AH® (Br(1)] = 30.7 kJ/mol - (OkJ/mol) =+30.7 kJ/mol - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculatethe entropy change of thesurroundingsasfollows: As a = -AH -30.7 kJ/mol 1000J 1kJ -30.7x10²J/mol

Finally, AS is calculatedas follows: AS war = 45° WX+ASUB 0 = 92.83J/K.mol - - 30.7x10²J/mol 30.7 x10 J/mol 92.83J/K.mol =330.7K - 273 =57.7°C Therefore, the boiling point of bromineis57.7°C.