At 298.15K, ΔG°f(HCOOH,g) = -351.0kJ mol-1 and ΔG°f(HCOOH,l) = -361.4kJ mol-1. Use R = 8.3145 J mol-1 K-1.
a) Calculate the vapor pressure of formic acid at 298.15 K.
Hint: treat the vaporization process as a reaction, with gas phase as the product, and use the relation between ln(Kp) and ΔG to find the vapor pressure.
Pvap,298.15 K = bar.
At 298.15K, ΔG°f(HCOOH,g) = -351.0kJ mol-1 and ΔG°f(HCOOH,l) = -361.4kJ mol-1. Use R = 8.3145 J...
Given that ΔH°f(HCOOH,g) = -378.7kJ mol-1 and ΔH°f(HCOOH,l) = -425.0kJ mol-1. Calculate the vapor pressure of formic acid in bar at 313.87 K, Using Clausius-clapeyron equation. Vapour pressure at 298.15K was calculated to be 0.0151bar from the previous question.
298.15K AH HO (1) -285.8 H2O(g) -241.8 4,6° Ch -237.1 -228.6 Sº Go Go 70.0 75.3 188.8 33.6 a) (1 pt) Calculate Acondensation Gº in J) for the condensation of water vapor at 25°C and 1 bar pressure. b) (1 pt) Is this process spontaneous under these conditions? c) (2 pts) Calculate Kp at 25°C. d) (15 pts) Use the Gibbs-Helmholtz equation AG d ) = -AH°(T) T 12 to calculate Acondensation Gº (in J) for the condensation of water...
1) Consider the equilibrium of methanol vapor and the liquid. CH3OH(l)↽−−⇀CH3OH(g) Thermodynamic Table at 25 ∘C Substance ΔHf∘ (kJ/mol) S∘ (J/mol‑K) ΔGf∘ (kJ/mol) CH3OH(l) −239.2 126.8 −166.6 CH3OH(g) −201.0 239.9 −162.3 What is the vapor pressure of the methanol at −30 ∘C? Pvap= atm What is the vapor pressure of the methanol at 40 ∘C? Pvap= 2) Substance ΔG°f(kJ/mol) M2O(s) −8.70 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represents a generic...
Please answer parts E-H and show
work/solution
ANSWER E-H
E F G H
ANSWER PARTS E, F, G, AND H........
8. We have two "named" equations which allow us to quantify the pressure- temperature relationship of a two-phase equilibrium, the Clausius equation, A, and the Clausius-Clapeyron equation, B din P ΔΗ dP AS A: dT RT2 The Clausius equation is an exact result for a one-component system, derived with no approximations, while the Clausius-Clapeyron equation has some constraints. We will...
Consider the reaction N2(g) + 3H2(g*2NH3(g) Using the standard thermodynamic data in the tables linked above, calculate ΔG n for this reaction at 298.15 K if the pressure of each gas is 43.08 mm Hg. ANSWER: kJ/mol Consider the reaction IH2(g) + C2H4(g)>C2Hf(g) Use the standard thermodynamic data in the tables linked above. Calculate Δ@for this reaction at 298.15K if the pressure of C2H6 g is reduced to 10.34 mm Hg, while the pressures of H2(g) and C2H4(g) remain at...
Review Constants Pe Consider the following two substances and their vapor pressures at 298 K All liquids evaporate to a certain extent. The pressure exerted by the gas phase in equilibrium with the liquid is called vapor pressure. Prap. The vapor pressure of a particular substance is determined by the strength of the intermolecular forces. But for any given substance, the vapor pressure only changes with temperature. The Clausius-Clapeyron equation expresses the relationship between vapor pressure and temperature: Substance Vapor...
1. Calculate the vapor pressure of heptane (b.p. 98.0 ∘C98.0 ∘C) at the gas chromatography column temperature of 94.0 ∘C94.0 ∘C using the form of the Clausius–Clapeyron equation shown ln(P1P2)=−(ΔHvapR)(1T1−1T2)ln(P1P2)=−(ΔHvapR)(1T1−1T2) where RR is the ideal gas constant, ΔHvapΔHvap is the enthalpy of vaporization, T1T1 and T2T2 are two different temperatures, and P1P1 and P2P2 are the vapor pressures at the respective temperatures. The enthalpy of vaporization can be estimated using Trouton's rule, ΔH∘vap=(88 J⋅mol−1⋅K−1)×Tbp.ΔHvap∘=(88 J·mol−1⋅K−1)×Tbp. P= 2. Calculate the vapor...
AHºf (kJ/mol) AG°f (kJ/mol) Sº (J/mol K) 0 0 130.7 Hydrogen H2 (g) H (g) H' (g) H+ (aq) 218.0 203.2 114.7 1536.2 0 -230.0 -157.0 -11.0 -285.8 -237.1 69.9 OH(aq) H20 (1) H20 (g) H2O2 (1) -241.8 -228.6 188.8 -187.8 -120.4 109.6 Iodine AH.(kJ/mol). | AGO. (k.I/mol) go (I/mol K | -53.0 -13.0 242.0 -277.7 -174.8 160.7 282.7 -235.1 -484.0 160.0 C2H40 (g, ethylene oxide) CH3CH2OH (1) CH3CH2OH (g) CH3COOH (1) C2H6 (g) C3H6 (g) CzH; (g) CH2=CHCN (1)...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
You have a 3.00-L container filled with N₂ (MM = 28.02 g/mol) at 298.15 K and 1.75 atm pressure connected to a 2.00-L container filled with Ar (MM = 39.95 g/mol) at 298.15 K and 2.15 atm pressure. A stopcock connecting the containers is opened and the gases are allowed to equilibrate between the two containers. What is the density of the final gas mixture? Assume ideal behavior. (Use R = 0.08206 L.atm/mol.K) (HINT: What is the total mass, m,...