1.000 mol of ammonia is vaporized at a constant pressure of 1.000 atm and a constant temperture of 240 K, the normal boiling temperture. The enthalpy change of the vaporization at this temperture is 25.11 kJ/mol. Find delta G, delta S, w and q.
1.000 mol of ammonia is vaporized at a constant pressure of 1.000 atm and a constant...
The vapor pressure of Y is 4.560 atm at 440K and 13.680 atm at 515K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(l) = Y(g). G440o = -5.55 kJ/mol G515o = -11.20 kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values delta S = J/(mol-K) delta H = kJ/mol What is the normal boiling point of Y in degrees Celcius?...
The vapor pressure of Y is 1.620 atm @ 200K and 4.698 atm @ 273K. What are the standard free energies of vaporization temperatures? Hint: vaporization reaction is Y(l) = Y(g) (at 200K) (at 273K) Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values. (Delta S degrees) (Delta H degrees) What is the normal boiling point of Y in degrees Celcius? t= ? degrees Celsius What is the vapor pressure...
The vapor pressure of Y is 4.590 atm at 330K and 9.180 atm at 380K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). AG330º = kJ/mol AG380º = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values ASO = J/(mol-K) AH = kJ/mol What is the normal boiling point of Y in degrees Celcius? t = Ос What...
The vapor pressure of Y is 4.050 atm at 290K and 13.365 atm at 373K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). «J/mol AG290º = AG373° = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values J/(mol-K) AS° = AH° = kJ/mol What is the normal boiling point of Y in degrees Celcius? 1 t= t =...
Ammonia has a normal boiling point of -33.3 °C. Under a pressure of 2.00 atm, what is ammonia’s boiling point in °C? For ammonia ∆H vaporization = 23.3 kJ/mol.
The vapor pressure of Y is 2.320 atm at 420K and 5.336 atm at 478K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(u) - Y(9). kJ/mol AG420º = AG478° - kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values 3/(mol-K) AS- AH kJ/mol What is the normal boiling point of Y in degrees Celclus? oc What is the vapor pressure...
Calculate the (equilibrium) sublimation temperature of CO_2 at a pressure of 2 atm assuming a constant enthalpy of sublimation of Delta H-26 kJ/mol. he sublimation temperature of CO_2 at 1 atm is -78.5 degree C. Gold has an equilibrium melting temperature of 1337 K at 1 atm and an enthalpy of fusion of 12, 550 J/mol. Find the pressure at which the equilibrium melting temperature is 1400 K. The densities of solid gold and liquid gold are 19.3 g/mL and...
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.
The normal boiling point of ammonia is -33.3°C, and its enthalpy of vaporization is 23.35 kJ/mol. Calculate the temperature in °C required to double the vapor pressure of ammonia.
6.34. The vapor pressure of benzene at 40.0°C is 0.241 atm. If the enthalpy of vaporization of C. Ha is 33.9 kJ/mol, estimate the normal boiling point of benzene.