17 Marks: 1 At 298 K, ?G o = - 6.36 kJ for the reaction: 2N 2 O(g) + 3O 2 (g) ?

Question

17 Marks: 1 At 298 K, ?Go = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ? 2N2O4(g)

Calculate ?G (in kJ) at 298 K when PN2O = 3.12 atm, PO2 = 0.0081 atm, and PN2O4= 0.515 atm.

Help given in feedback.
Answer: Question18 Marks: 1 Under the conditions in the previous question, in which direction is the reaction spontaneous? Choose one answer. a. to the left (more reactant) b. to the rignt (more product) Question19 Marks: 1 At 298 K, ?Go = + 8.68 kJ for the reaction:

ZnF2(s) ? Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ?G means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.) Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Question20 Marks: 1 Calculate ?G (in kJ) at 298 K for some solid ZnF2, 0.011 M Zn2+ and 0.058 M F-(aq).
Hint given in feedback.
Answer: Question21 Marks: 1
Under the conditions of the previous question what is the spontaneous direction? Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Question22 Marks: 1 Endothermic reaction; decrease in entropy:
Calculate the equilibrium constant at 34 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Hint given in feedback.
Answer: Question23 Marks: 1 Calculate the equilibrium constant at 126 K for the thermodynamic data in the previous question.

Notice that Keq is larger at the larger temperature for an endothermic reaction.
Answer: Question24 Marks: 1 Endothermic reaction; increase in entropy
Calculate the equilibrium constant at 13 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = 100 J/K.
Answer: Question25 Marks: 1 Calculate the equilibrium constant at 130 K for the thermodynamic data in the previous question.

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.
Answer: Question26 Marks: 1 Warm-up question Using data from the Appendix, calculate ?Go (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ? N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions) Answer: Question27 Marks: 1 Calculate ?G (in kJ) at 65 oC for the reaction:

N2O(0.0029 atm) + H2(0.18 atm) ? N2(95.2 atm) +H2O(l) Answer: Question28 Marks: 1 Assume that the ?Ho and ?So of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 57 oC (in atm).

CH3OH (l) ? CH3OH(g) . . . ?Ho = 38.0 kJ and ?So = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback. Answer: Question29 Marks: 1 Calculate the vapor pressure of Hg at 50 oC (in atm).

Hg(l) ? Hg(g) . . . ?Ho = 61.32 kJ and ?So = 98.83 J/K Answer: Question30 Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C 17 Marks: 1 At 298 K, ?Go = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ? 2N2O4(g)

Calculate ?G (in kJ) at 298 K when PN2O = 3.12 atm, PO2 = 0.0081 atm, and PN2O4= 0.515 atm.

Help given in feedback.
Answer: 17 Marks: 1 Marks: 1 At 298 K, ?Go = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ? 2N2O4(g)

Calculate ?G (in kJ) at 298 K when PN2O = 3.12 atm, PO2 = 0.0081 atm, and PN2O4= 0.515 atm.

Help given in feedback.
Answer: At 298 K, ?Go = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ? 2N2O4(g)

Calculate ?G (in kJ) at 298 K when PN2O = 3.12 atm, PO2 = 0.0081 atm, and PN2O4= 0.515 atm.

Help given in feedback.
Answer: Answer: Question18 Marks: 1 Under the conditions in the previous question, in which direction is the reaction spontaneous? Choose one answer. a. to the left (more reactant) b. to the rignt (more product) Question18 Marks: 1 Marks: 1 Under the conditions in the previous question, in which direction is the reaction spontaneous? Choose one answer. a. to the left (more reactant) b. to the rignt (more product) Under the conditions in the previous question, in which direction is the reaction spontaneous? Choose one answer. a. to the left (more reactant) b. to the rignt (more product) Choose one answer. Question19 Marks: 1 At 298 K, ?Go = + 8.68 kJ for the reaction:

ZnF2(s) ? Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ?G means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.) Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Question19 Marks: 1 Marks: 1 At 298 K, ?Go = + 8.68 kJ for the reaction:

ZnF2(s) ? Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ?G means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.) Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) At 298 K, ?Go = + 8.68 kJ for the reaction:

ZnF2(s) ? Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ?G means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.) Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Choose one answer. Question20 Marks: 1 Calculate ?G (in kJ) at 298 K for some solid ZnF2, 0.011 M Zn2+ and 0.058 M F-(aq).
Hint given in feedback.
Answer: Question20 Marks: 1 Marks: 1 Calculate ?G (in kJ) at 298 K for some solid ZnF2, 0.011 M Zn2+ and 0.058 M F-(aq).
Hint given in feedback.
Answer: Calculate ?G (in kJ) at 298 K for some solid ZnF2, 0.011 M Zn2+ and 0.058 M F-(aq).
Hint given in feedback.
Answer: Answer: Question21 Marks: 1
Under the conditions of the previous question what is the spontaneous direction? Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Question21 Marks: 1 Marks: 1
Under the conditions of the previous question what is the spontaneous direction? Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves)
Under the conditions of the previous question what is the spontaneous direction? Choose one answer. a. to the left (more precipitate forms) b. to the right (more solid dissolves) Choose one answer. Question22 Marks: 1 Endothermic reaction; decrease in entropy:
Calculate the equilibrium constant at 34 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Hint given in feedback.
Answer: Question22 Marks: 1 Marks: 1 Endothermic reaction; decrease in entropy:
Calculate the equilibrium constant at 34 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Hint given in feedback.
Answer: Endothermic reaction; decrease in entropy:
Calculate the equilibrium constant at 34 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Hint given in feedback.
Answer: Answer: Question23 Marks: 1 Calculate the equilibrium constant at 126 K for the thermodynamic data in the previous question.

Notice that Keq is larger at the larger temperature for an endothermic reaction.
Answer: Question23 Marks: 1 Marks: 1 Calculate the equilibrium constant at 126 K for the thermodynamic data in the previous question.

Notice that Keq is larger at the larger temperature for an endothermic reaction.
Answer: Calculate the equilibrium constant at 126 K for the thermodynamic data in the previous question.

Notice that Keq is larger at the larger temperature for an endothermic reaction.
Answer: Answer: Question24 Marks: 1 Endothermic reaction; increase in entropy
Calculate the equilibrium constant at 13 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = 100 J/K.
Answer: Question24 Marks: 1 Marks: 1 Endothermic reaction; increase in entropy
Calculate the equilibrium constant at 13 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = 100 J/K.
Answer: Endothermic reaction; increase in entropy
Calculate the equilibrium constant at 13 K for a reaction with ?Hrxno = 10 kJ and ?Srxno = 100 J/K.
Answer: Answer: Question25 Marks: 1 Calculate the equilibrium constant at 130 K for the thermodynamic data in the previous question.

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.
Answer: Question25 Marks: 1 Marks: 1 Calculate the equilibrium constant at 130 K for the thermodynamic data in the previous question.

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.
Answer: Calculate the equilibrium constant at 130 K for the thermodynamic data in the previous question.

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.
Answer: Answer: Question26 Marks: 1 Warm-up question Using data from the Appendix, calculate ?Go (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ? N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions) Answer: Question26 Marks: 1 Marks: 1 Warm-up question Using data from the Appendix, calculate ?Go (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ? N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions) Answer: Warm-up question Using data from the Appendix, calculate ?Go (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ? N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions) Answer: Answer: Question27 Marks: 1 Calculate ?G (in kJ) at 65 oC for the reaction:

N2O(0.0029 atm) + H2(0.18 atm) ? N2(95.2 atm) +H2O(l) Answer: Question27 Marks: 1 Marks: 1 Calculate ?G (in kJ) at 65 oC for the reaction:

N2O(0.0029 atm) + H2(0.18 atm) ? N2(95.2 atm) +H2O(l) Answer: Calculate ?G (in kJ) at 65 oC for the reaction:

N2O(0.0029 atm) + H2(0.18 atm) ? N2(95.2 atm) +H2O(l) Answer: Answer: Question28 Marks: 1 Assume that the ?Ho and ?So of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 57 oC (in atm).

CH3OH (l) ? CH3OH(g) . . . ?Ho = 38.0 kJ and ?So = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback. Answer: Question28 Marks: 1 Marks: 1 Assume that the ?Ho and ?So of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 57 oC (in atm).

CH3OH (l) ? CH3OH(g) . . . ?Ho = 38.0 kJ and ?So = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback. Answer: Assume that the ?Ho and ?So of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 57 oC (in atm).

CH3OH (l) ? CH3OH(g) . . . ?Ho = 38.0 kJ and ?So = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback. Answer: Answer: Question29 Marks: 1 Calculate the vapor pressure of Hg at 50 oC (in atm).

Hg(l) ? Hg(g) . . . ?Ho = 61.32 kJ and ?So = 98.83 J/K Answer: Question29 Marks: 1 Marks: 1 Calculate the vapor pressure of Hg at 50 oC (in atm).

Hg(l) ? Hg(g) . . . ?Ho = 61.32 kJ and ?So = 98.83 J/K Answer: Calculate the vapor pressure of Hg at 50 oC (in atm).

Hg(l) ? Hg(g) . . . ?Ho = 61.32 kJ and ?So = 98.83 J/K Answer: Answer: Question30 Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C Question30 Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C a. to the left (more reactant) b. to the rignt (more product)

Explanation / Answer

17)

delta G = delta Go +RTlnQ = -6.36 KJ + 8.314*298*ln [ (0.515)^2 *10^-15/ (3.12)^2 *(0.081)^3]. =-6360 - 75817.9 = -82177 J = -82.177 KJ.

18)

since delta G is negative the reaction proceeds towards the products.

19)

In this delta Go is positive and also RTlnQ is is zero as ln1 = 0.

so the total G is positive and the reaction proceeds towards left, towards the reactant.

20)

delat G = delta Go +RTln [ (F-)^2 * (Zn2+) ]

= 8.68 + 8.314*298*ln [0.058^2 *0.011]

= 8680 - 25282

= -16602.34 J

21)

since the total G is negative the reaction procees to the left (more precipitate forms).

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