The following are thermodynamic data for isopropanol, an alcohol
commonly known as rubbing alcohol: ΔH°f = -317.0 kJ mol-1 ΔH°comb =
-127.2 kJ mol-1 ΔH fus = 5.3 kJ mol-1 ΔH vap = 44.0 kJ mol-1 S°m =
180.6 J mol-1 K-1 ΔS fus = 28.6 J mol-1 K-1 ΔS vap = 123.9 J mol-1
K-1 (a) At the boiling point, liquid and vapor phase of isopropanol
are at equilibrium (ΔG = 0). Choose appropriate thermodynamic data
to calculate the boiling point of isopropanol. (b) Compare your
calculated value with the value reported in the literature (or
online).
A) TBP =∆H°vap/∆S°vap
= 44.0×1000/123.9
= 355 K => 82.1 °c
B) Value reported from literature = 82.5 °C
So, values are approximately equal as calculated and online data
The following are thermodynamic data for isopropanol, an alcohol commonly known as rubbing alcohol: ΔH°f =...
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 185.98 kJ/mol ΔS∘rxn 27.40 J/(mol⋅K) Calculate the temperature in Kelvin above which this reaction is spontaneous. Express your answer to 0 decimal places and in K.
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 179.97 kJ/mol ΔS∘rxn 22.99 J/(mol⋅K) Calculate the equilibrium constant for the following reaction at room temperature, 295.9 K: Whatever answer you get multiply by 1x1031 and enter that number to 4 decimal places.
Use the thermodynamic data provided below to estimate the boiling point (in K) of V. Report your answer to zero decimal places in standard notation (i.e. 123. kJ *For numbers ending in zero, be sure to include the decimal!*). Substance ΔH°f (kJ/mol) S° (J mol-1K-1) V (l) 17.3 36.1 V (g) 515.5 182.3
Use the thermodynamic data provided below to estimate the boiling point (in K) of CCl4. Report your answer to zero decimal places in standard notation (i.e. 123. kJ *For numbers ending in zero, be sure to include the decimal!*). Substance ΔH°f (kJ/mol) S° (J mol-1K-1) CCl4 (l) -128.4 214.4 CCl4 (g) -96 309.6
1a) Consider the following reaction: 3 C(s) + 4 H2(g) → C3H8(g); ΔH° = –104.7 kJ; ΔS° = –287.4 J/K at 298 K What is the equilibrium constant at 298 K for this reaction? Report answer to TWO significant figures. 1b) Τhe enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33 °C). Calculate the value of ΔSsurr when 1.00 mole of ammonia is vaporized at –33 °C and 1.00 atm. Report answer to THREE significant figures.
Given the following thermodynamic data calculate ΔS and ΔSsurr for the following reaction at 25°C and 1 atm. CH4g)+ 2028) CO2g) +2H20(g) So (J/K.mol) 186 205 214 189 ΔΙ!of (kJ/mol) CH4g) 75 02(g) CO2(8) -394 H20(g) -242 J/K J/K Submit Show Hints
Consider the data in the table. Compound Melting point (°C) AHfus (kJ/mol) Boiling point (°C) AH ap (kJ/mol) HF -83.11 4.577 19.54 25.18 НСІ -114.3 1.991 -84.9 17.53 HBr - 86.96 2.406 -67.0 19.27 HI -50.91 2.871 -35.38 21.16 Using the data in the table, calculate ASfus and ASvap for HI. ASfus = J/(K.mol) AS vap = J/(K · mol) Determine the entropy change when 6.50 mol HI(1) boils at atmospheric pressure. AS = J/K
6. Use the following thermodynamic data to calculate me following thermodynamic data to calculate AS for the formation of Fe2O3 at 298.2 K Species Fe(s) O2(g) Fe2O3(s) AH (kJ/mol-rxn) S (J/Komol-rxn) AH producers - 0.0 27.8 0.0 Alveolu 205.1 -825.5 87.4 -825.5 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) a) -2202.7J/ KX b) -546.2 J/K AS-A Sproducts-AS rearsants c) +546.2 J/K d) +2202.7J/ KV (2 X 87,4)-((4X27.8)+(3x205.1)) e) +4984.9 J/K 1 = 174,8-(111.27615.3) Ssurr -4H = 174,8-726.5 Assus =-551.75l...
Consider the data in the table. Compound Melting point (°C) AH (kJ/mol) Boiling point (°C) AHap (kJ/mol HF -83.11 4.577 19.54 25.18 HCI -114.3 1.991 -84.9 17.53 HBr -86.96 -67.0 19.27 HI -50.91 -50.91 2.871 -35.38 21.16 Using the data in the table, calculate ASfus and AS vap for HI. 2.406 A Stus = 24.1 J/(K mol) ASvap = J/(K.mol) Determine the entropy change when 7.50 mol HI(1) freezes at atmospheric pressure. AS = J/K Question Source: MRG - General...
MAIN QUESTIO Given the following thermodynamic data calculate AS for the following reaction at 25°C and 1 atm: XeF(9) —— XeFe(s) + Fz(9) AH®: (kJ/mol) S (J/K-mol) XeF6(9) -294 300. XeF(s) -251 146 F2(g) 10 203