From the information, determine the ratio of
pressures. Hvap of 32.1 kJ/mol, 35°C and 117°C
options:
A) 7.29×10^32
B) 13.8
C) cannot calculate
D) 0.072
E) 1.37×10^-33
According to Clapeyron equation :
ln (P2/P1) = -Hvap/R
(1/T2-1/T1)
T1 = 35 +273.15 = 308.15 K
T2 = 117 +273.15 = 390.15 K
Hvap
= 32.1 kJ/mol = 32.1 x 1000 = 32100 J/mol
R = 8.314 J/mol K
ln (P2/P1) = - 32100/8.314 ((1/390.15)-(1/308.15))
ln (P2/P1) = -3860.96 ((0.00256 - 0.00324))
ln (P2/P1) = -3860.96 ((-0.00068)
ln (P2/P1) = 2.625
(P2/P1) = e2.625 = 13.80
P2/P1 = 13.80
P1/P2 = 1/13.80 = 0.072
From the information, determine the ratio of pressures. Hvap of 32.1 kJ/mol, 35°C and 117°C options:...
Chloroform has a normal boiling point of 61°C and a Hvap of 29.6 kJ/mol. Determine the vapor pressure (in atm) of chloroform at 35°C. Give you answer to decimal places
Cyanogen (C2N2 Hvap=23.3 kJ/mol) has a normal boiling point of -21.2C. At what temperature in C, will the vapor pressure of liquid cyanogen be 0.1000atm? a. -294 b. -21.1 c. 44.6 d. -272 e. -64.3
Consider the following two reactions: A → 2B ΔH°rxn = 456.7 kJ/mol A → C ΔH°rxn = -22.1 kJ/mol Determine the enthalpy change for the process: 2B → C Question 40 options: A) 434.6 kJ/mol B) -434.6 kJ/mol C) -478.8 kJ/mol D) 478.8 kJ/mol E) More information is needed to solve the problem.
31. If the AG values for reactions A and B are -100 kj/mol and -10 kJ/mol, then: a. Reaction A proceeds at a faster rate than reaction B b. Reaction B proceeds at a faster rate than reaction A C. Only reaction A is spontaneous d. Only reaction B is spontaneous e. No conclusion can be deduced regarding the rates of the reactions 32. On the following plot, the estimated Vmax for the enzyme is: a. -0.02 b. 25 C....
Determine the free energy (in kJ/mol) for the decomposition of ammonia at 25.0 °C when the partial pressures are PN2 = 0.87 atm, PH2 = 0.25 atm and PNH3 = 12.9 atm, and ΔG°rxn -1 =33.0kJmol . Calculate the vapor pressure (in Torr) of mercury under standard conditions. Use Appendix D to look up the required standard thermodynamic values. (Recall the pressure conversions: 1 atm = 1.01325 bar = 1.01325x105 Pa = 760 Torr) Benzene has a vapor pressure of...
At 25°C, the following heats of reaction are known: AH (kJ/mol 167.4 2CIF + 02 →Cl20 + F20 2ClF3 + 202 →Cl20+3F20 341.4 2F2 + 02 → 2F20 At the same temperature, calculate ΔH for the reaction: ClF + F2 → CIF3 -43.4 A. -217.5 kJ/mol B.-130.2 kJ/mol C. +217.5 kJ/mol ○ D.-108.7 kJ/mol E. none of these QUESTION 4 Consider the reaction: When a 12.9-g sample of ethyl alcohol(molar mass 46.07 g/mol) is burned, how much energy is released...
10. Given the following table: Compound NO; (g) N:04 (g) AHP (kJ mol') 33.85 9.66 AGP (kJ mol') 51.84 98.29 For the reaction: N;04 (g) 2NO, (g) (a) Use the information in the Table to calculate AGⓇ for the reaction (b) Use the information in the Table to calculate AH® for the reaction (c) Calculate K, at 25°C. (d) Calculate Kp at 1600 °C. (Assume AH to be temperature independent) (e) Calculate the degree of dissociation, c., of N2O4 at...
Question 19 + 0,(2) ► 4 CO,(g)+5 H20(8) given the following information. AG, (kJ/mol) -15.0 -394.4 -228.57 Determine AG rxn for C, H10 Substance C4H101) CO2(g) H 2018) a.-1791.0 kJ/mol O b.-2735.5 kJ/mol O c. -2705.5 kJ/mol d. +608.0 kJ/mol e. -608.0 kJ/mol
The enthalpy of combustion (ΔH°c) of 1,1,2,2,-tetramethylcyclopropane (C7H14) is -4635.62 kJ/mol. a-Using the appropriate information given below, calculate the enthalpy of formation (ΔH°f), in kJ/mol, for 1,1,2,2,-tetramethylcyclopropane. Report your answer to two decimal places. ΔH°f (CO2 (g)) = -393.51 kJ/mol ΔH°f (H2O (l)) = -285.83 b- Determine the mass (in g) of 1,1,2,2,-tetramethylcyclopropane produced, if ΔH° was determined to be -35.93 kJ during an experiment in which 1,1,2,2,-tetramethylcyclopropane was formed. Report your answer to three significant figures.
19. Use the AHºf information provided to calculate AHºrxn for the following: 10 pts AHºf (kJ/mol) SO2Cl2(g) + 2 H20(1) ► 2 HC (g) + H2SO4(1) AH°rxn- ? SO2Cl2(g)-364 H20(1) HCl(g) -286 1(-364) + 2 (-286) -2 (-92) + 1 (-814) = -364-572 +184 - 814 -92 H2SO4(1)-814 A) -256 kJ B) +161 kJ C) -62 kJ D) +800. kJ E) -422 kJ