B. Correct! Good work. F = C P + 2 = = 2 degrees of freedom. Good try. Hint: Think about the meaning of components and phases.

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1 Physical Chemistry - Problem Drill 06: Phase Equilibrium No. 1 of The Gibbs Phase Rule is F = C P + 2, how many degrees of freedom does a system have that has two independent components and two phases? (A) 1 (B) 2 (C) 3 (D) 4 (E) None of the above Good try. Hint: Think about the meaning of components and phases. B. Correct! Good work. F = C P + 2 = = 2 degrees of freedom. Good try. Hint: Think about the meaning of components and phases. Good try. Hint: Think about the meaning of components and phases. Good try. Hint: Think about the meaning of components and phases. Correct Answer: B F = C P + 2 = = 2 degrees of freedom.

2 No. 2 of A two-phase region between phase composition lines is a region with one phase of compositions x a and another phase of compositions y a identified by. (A) Vertical tie lines. (B) Vertical phase lines. (C) Horizontal phase lines. (D) Horizontal tie lines. (E) None of the above. Good try. Hint: Think about the meaning of phase composition lines. Good try. Hint: Think about the meaning of phase composition lines. Good try. Hint: Think about the meaning of phase composition lines. D. Correct! Good work. A two-phase region between phase composition lines is a region with one phase of compositions x a and another phase of compositions y a identified by horizontal tie lines. Good try. Hint: Think about the meaning of phase composition lines. Correct Answer: D A two-phase region between phase composition lines is a region with one phase of compositions x a and another phase of compositions y a identified by horizontal tie lines.

3 No. 3 of phases may coexist at a Eutectic Point. (A) 4 (B) 3 (C) 2 (D) 1 (E) None of the above Good try. Hint: Think about the meaning of a Eutectic Point. B. Correct! Good work. Three phases may coexist at a Eutectic Point. Good try. Hint: Think about the meaning of a Eutectic Point. Good try. Hint: Think about the meaning of a Eutectic Point. Good try. Hint: Think about the meaning of a Eutectic Point. Correct Answer: B Three phases may coexist at a Eutectic Point.

4 No. 4 of Ethanol has a boiling point of 79 at 1 atm and standard enthalpy of change of vaporization (Δ vap H o ) of kj/mol. Calculate its vapor pressure at 60. (A) 0 (B) kpa (C) kpa (D) 1 atm (E) Pa Good try. The vapor pressure cannot be zero since there is ethanol presented in the vapor. Hint: Use the Clausius-Clapeyron equation to calculate the vapor pressure. B. Correct! Good job! The Clausius-Clapeyron equation is used to calculate the vapor pressure. Good try. The calculation is correct but with the wrong sign. Good try. The partial pressure should be less than 1 atm because of lower temperature. Hint: Use the Clausius-Clapeyron equation to calculate the vapor pressure. Good try. Check the calculation. Hint: Use the Clausius-Clapeyron equation to calculate the vapor pressure. Correct Answer: B The Clausius-Clapeyron equation calculates a second point on a liquid-vapor or solid-vapor line when the heat of vaporization and one point on the line are known. p2 Δ T vaporization H 2 dt Δvaporization H 1 1 d ln p ln p2 ln p1 -. p R T 2 T R T 2 T lnp 2 lnp 1 = - (Δ vap H o /R)(1/T 1/T b ) lnp 2 ln( x10 5 Pa) = -[(38.56x1000 J/mol)/(8.314J/K*mol)]x[(1/333.2K 1/352.2K) = p 2 = 47,819 Pa = kpa

5 No. 5 of Pressure can alter melting point. Calculate the pressure change needed to alter Mercury s melting temperature by 1 C. It has a melting point of C, and the density of Hg(l) = 13.7 g/ml and Hg(s) = 14.2 g/ml. The enthalpy of phase change is 9.75 J/g. (A) 1atm (B) 0 (C) 1.62 x 10 4 kpa (D) 3.25 kpa (E) 12 Pa Good try. Hint: Use the Clapeyron Equation and density formula. Be aware of all the unit conversion. Good try. Hint: Use the Clapeyron Equation and density formula. Be aware of all the unit conversion.. C. Correct! Good Job! Use the Clapeyron Equation and density formula to get the pressure changed needed to raise 1K or 1 C. Good try. Hint: Use the Clapeyron Equation and density formula. Be aware of all the unit conversion. Good try. Hint: Use the Clapeyron Equation and density formula. Be aware of all the unit conversion. Correct Answer: C Strategy: This is the process of solid-liquid phase transition. The Clapeyron Equation can be used to relate the pressure change to temperature change. dp fusionh dt T V fusion With fusion H and T given, we just need to figure out fusion V. Giving 1 gram of Mercury and Volume = Mass/Density, we can calculate the fusion V: fusion V = V(liquid) V(solid) = (1g /13.7 g*ml -1 ) (1g/14.2 g*ml) = 2.57 x10-3 ml We can change dp/dt to p/ T by assuming a straight line on the pressuretemperature phase diagram in the mercury solid-liquid equilibrium. Plus all known values into the Claperyron Equation with J = kg*m 2 *s -2 and Pa = kg*m -1 s -2. Pa = J*m -3 p/ T = (9.75 J/g) /[(234.3K)x(2.57x10-3 ml/g x 10-6 m 3 /ml)] = 1.6 x10 7 J*m -3 /K = 1.62 x10 7 Pa/K = 16.2 x 10 3 kpa/k p = (16.2 kpa/k) x T = (16.2 x 10 3 kpa/k) x 1K =1.62 x 10 4 kpa

6 No. 6 of Distillation of a solution having either a high-boiling or a low-boiling azeotrope of components A and B ultimately leads to constant boiling at the boiling point of. (A) Pure component A. (B) Pure component B. (C) The azeotrope of A and B. (D) A compound of A and B. (E) None of the above. Good try. Hint: Think about the properties of an azeotrope. Good try. Hint: Think about the properties of an azeotrope. C. Correct! Good work. Distillation of a solution having either a high-boiling or a low-boiling azeotrope of components A and B ultimately leads to constant boiling at the boiling point of the azeotrope. Good try. Hint: Think about the properties of an azeotrope. Good try. Hint: Think about the properties of an azeotrope. Answer: C Distillation of a solution having either a high-boiling or a low-boiling azeotrope of components A and B ultimately leads to constant boiling at the boiling point of the azeotrope.

7 No. 7 of The relationship n x L x = n y L y is the and involves distances along a. A. Phase Rule. Tie Line B. Lever Rule. Phase Line C. Phase Rule. Phase Line D. Lever Rule. Tie Line. E. None of the above Good try. Hint: Think about phase equilibrium. Good try. Hint: Think about phase equilibrium. Good try. Hint: Think about phase equilibrium. D. Correct! Good work. It is indeed the Lever Rule. Good try. Hint: Think about phase equilibrium. Answer: D The relationship n x L x = n y L y is the Lever rule and involves distances along a Tie Line.

8 No. 8 of Calculate the vapor pressure of liquid Br 2 at standard state, giving f G = 3.11 kj/mol of the phase equilibrium Br 2 (l) Br 2 (g). (A) 0 (B) 1 bar (C) bar (D) bar (E) 1.23 bar The vapor pressure cannot be zero since the gas-liquid phase is in its equilibrium where there is a gas vapor of Br 2. Hint: Use the Gibb s free energy equation f G = - RT ln (K p ) and equilibrium constant K p of the phase transition Br 2 (l) Br 2 (g). Good try. The vapor pressure should be smaller than the standard state pressure. Hint: Use the Gibb s free energy equation f G = -RT ln (K p ) and equilibrium constant K p of the phase transition Br 2 (l) Br 2 (g). Good try. The temperature should be 298K not 273K. Hint: Use the Gibb s free energy equation f G = -RT ln (K p ) and equilibrium constant K p of the phase transition Br 2 (l) Br 2 (g). D. Correct! Good job! Use the Gibb s free energy equation f G = -RT ln (K p ) and equilibrium constant K p of the phase transition Br 2 (l) Br 2 (g). Good try. Check your algebra. Hint: Use the Gibb s free energy equation f G = -RT ln (K p ) and equilibrium constant K p of the phase transition Br 2 (l) Br 2 (g). Correct Answer: D Strategy For a phase transition Br 2 (l) Br 2 (g), one can relate K p to f G via f G = -RT ln (K p ). The vapor pressure can then be calculated from the equilibrium constant K p = P Br2(g) /P o. Standard state (1bar and 298K) is different STP. Let s plug all the data in with the proper units: 3.11 x 10 3 J/mol = -(8.314 J/K) (298K) ln (P Br2(g) /1bar) Therefore, P Br2(g) = bar

9 No. 9 of For a two-component system, what is the maximum number of phases it could be in equilibrium? A. 0 B. 1 C. 2 D. 3 E. 4 Good try. Hint: To get the maximum phases, set the degrees of freedom to zero, F = C + P + 2. Good try. Hint: To get the maximum phases, set the degrees of freedom to zero, F = C + P + 2. Good try. Hint: To get the maximum phases, set the degrees of freedom to zero, F = C + P + 2. Good try. Hint: To get the maximum phases, set the degrees of freedom to zero, F = C + P + 2. E. Correct! Good job. To get the maximum phases, set the degrees of freedom to zero, F = C - P + 2, that is, 0 = 2 - P max + 2, thus P max = 4. Correct Answer: E According to the Gibb s Phase Rule F = C + P + 2, if C is fixed, maximum phases cab be obtained if F is at its minimum. To get the maximum phases, set the degrees of freedom to zero, F = C - P + 2, that is, 0 = 2 - P max + 2, thus P max = 4.

10 No. 10 of The Degrees of Freedom for a system F is the number of independently variable variables for the system. (A) Mechanical. (B) Thermodynamic. (C) Extensive. (D) Intensive. (E) None of the above. Good try. Hint: Think about the Gibbs Phase Rule. Good try. Hint: Think about the Gibbs Phase Rule. Good try. Hint: Think about the Gibbs Phase Rule. D. Correct! Good work. The Degrees of Freedom for a system F is the number of independently variable-intensive variables for the system. Good try. Hint: Think about the Gibbs Phase Rule. Correct Answer: D The Degrees of Freedom for a system F is the number of independently variable-intensive variables for the system.

There are five problems on the exam. Do all of the problems. Show your work

CHM 3400 Fundamentals of Physical Chemistry Second Hour Exam March 8, 2017 There are five problems on the exam. Do all of the problems. Show your work R = 0.08206 L atm/mole K N A = 6.022 x 10 23 R = 0.08314