The normal boiling point of C_2CI_3F_3 is 47.6 degreeC and its molar enthalpy of

Question

The normal boiling point of C_2CI_3F_3 is 47.6 degreeC and its molar enthalpy of vaporization is 27.49 kJ/moL What is the change in entropy in the system in J/K when 28.6 grams of C_2CI_3F_3 vaporizes to a gas at the normal boiling point? A) 13.1 B) -13.1 C) 4.19 D) 27.5 E) -4.19 5) The value of DeltaGdegree (kJ/mol) at 25 degreeC is -57.5. What is the equilibrium constant for this reaction at this temperature expressed with 2 significant digits? A) 8.4 times 10^101 B) 10 C) 1.2 times 10^10 D) 1.0 E) more information is needed 6) For a given reaction with DeltaS = -50.8 J/K-mol, the DeltaG = 0 at 395 K. The value of DeltaH must be kJ/mol, assuming that DeltaH and DeltaS do not vary with temperature. A) 20.1 B) -20.1 C) -7.78 times 10^-3 D) 1.29 times 10^-4 E) -1.29 times 10^-4 7) For an isothermal process, DeltaS =. A) q + w B) q C) T_qrev D) qrev E) q_rev/T

Explanation / Answer

4) Given that Hvap = 27.49 kJ/mol = 27490 J/mol

   Moles of C2Cl3F3 = mass / molar mass = 28.6 g / 187.5 g/mol = 0.1525 mol

   T = 47.6 oC = 47.6 +273 K = 320.6 K

S = nH/T

   =  0.1525 mol x 27490 J/mol / 320.6 K

   = 13.1 J/K

Therefore,

   change in entropy = 13.1 J/K

5) Given that

Gorxn = -57.5 kJ/mol = - 57500 J/mol

T = 25oC = 298 K

Gorxn = -RT ln K where R = 8.314 J/mol/K

  Kp = e (- Gorxn/RT)  

= e -( -57500/ 8.314 x 298)

= 1.2 X 1010

Therefore,

K = 1.2 X 1010

6)   G = H -  T S

If  G = 0,

then H -  T S = 0

  H = T S

= 395 x -50.8 J/K

= -20.1 kJ/mol

Therefore,

H = -20.1 kJ/mol

7) For isothermal process,

S = qrev/T

  

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.