Name:___________________________



Name:______[KEY]_______________________ Date:________________ Period:____

[KEY] WS Thermodynamics [KEY]

SECTION II: Free Response (Calculator Permitted)

CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS. It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do not. Attention should be paid to significant figures

2 NO(g) + O2(g) ( 2 NO2(g) (H°= –114.1 kJ/molrxn, (S°= –146.5 J/molrxn·K

1. The reaction represented above is one that contributes significantly to the formation of photochemical smog.

(a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g). (2)

73.1 g NO x 1 mol NO x 114.1 kJ = 139 kJ

30.01 g NO 2 mol NO

(b) For the reaction at 25(C, the value of the standard free-energy change, (G(, is

–70.4 kJ/molrxn.

(i) Calculate the value of the equilibrium constant, Keq, for the reaction at 25(C. (2)

K = e −ΔG/RT

K = e –(–70.4 kJ)/(0.008314 kJ mol-1 K-1) (298 K)

K = 2.19 x 1012

(ii) Indicate whether the value of (G( would become more negative, less negative, or

remain unchanged as the temperature is increased. Justify your answer. (2)

ΔG° = ΔH° – TΔS°

(–) = (–) – T (–)

(–) = (–) (+)

As the temperature increases, the positive entropy term (−TΔS°) will become larger making ΔG° become less negative.

(c) Use the data in the table below to calculate the value of the standard molar entropy, S(,

for O2(g) at 25(C. (2)

| |Standard Molar Entropy, S( (J mol–1·K–1)|

|NO(g) |210.8 |

|NO2(g) |240.1 |

ΔS orxn = ΣΔS f products – ΣΔS f reactants

ΔS orxn = 2( S NO2 ) − [2(S NO) + SO2 ]

– 146.5 = 2 (240.1) – [2(210.8) + SO2]

SO2 = 205.1 J mol–1∙K–1

(d) Use the data in the table below to calculate the bond energy, in kJ mol–1, of the nitrogen-oxygen bond in NO2 . Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy). (2)

| |Bond Energy |

| |(kJ∙mol–1) |

|Nitrogen-oxygen bond in NO |607 |

|Oxygen-oxygen bond in O2 |495 |

|Nitrogen-oxygen bond in NO2 |? |

∆H = Σ(B.E.)reactants – Σ(B.E.)products

ΔHrxn = [2(N≡O) + O=O] − 4[N=O]

−114.1 = [2(607) + 495] − 4 (x)

N=O bond energy = 456 kJ mol–1

2. Nitrogen monoxide, NO(g), and carbon monoxide, CO(g), are air pollutants generated by automobiles. It has been proposed that under suitable conditions these two gases could react to form N2(g) and CO2(g), which are components of unpolluted air.

(a) Write a balanced equation for the reaction described above. (1)

2 NO(g) + 2 CO(g) → N2(g) + 2 CO2(g)

(b) Write the expression for the equilibrium constant, Kp , for the reaction. (1)

Kp = (PN2)(PCO2)2

(PNO)2(PCO)2

(c) Consider the following thermodynamic data.

NO CO CO2 .

∆Gfo (kJ mol−1) +86.55 −137.15 −394.36

(i) Calculate the value of ΔG° for the reaction at 298 K. (2)

ΔG°rxn = ΣΔGf° (products) – ΣΔGf° (reactants)

= 2 ΔGf° (CO2) – [2 ΔGf° (NO) + 2 ΔGf° (CO)]

= 2(−394.36) – [2(86.55) + 2(–137.15)]

= –687.52 kJ

(ii) Given that ΔH° for the reaction at 298 K is −746 kJ per mole of N2(g) formed,

calculate the value of ΔS° for the reaction at 298 K. Include units with answer. (2)

ΔG°rxn = ΔH°rxn – TΔS°rxn

–687.52 kJ = (–746 kJ) – (298) ΔS°rxn

ΔS°rxn = – 0.196 kJ∙mol–1∙K–1 or –196 J∙mol–1∙K–1

(d) For the reaction at 298 K, the value of Kp is 3.3 × 10120. In an urban area, typical

pressures of the gases in the reaction are PNO = 5.0 × 10−7 atm, PCO = 5.0 × 10−5 atm,

PN2 = 0.781 atm, and PCO2 = 3.1 × 10–4 atm.

(i) In which direction (to the right or to the left) will the reaction be favorable

at 298 K with these partial pressures? Explain. (2)

Q = (PN2)(PCO2)2 = (0.781)(3.1 x 10–4)2 = 1.2 x 1014

(PNO)2(PCO)2 (5.0 x 10–7)2(5.0 x 10–5)2

Q < K

The reaction will proceed to the right to reach equilibrium.

(ii) The value of ΔG for the reaction at 298 K when the gases are at the partial pressures

given above is expected to be greater than 0, less than 0, or equal to 0. (1)

The reaction is favorable in the forward direction b/c Q < K, therefore ΔG must be negative in the forward direction.

(ΔG° should NOT be discussed)

Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.

3. Answer the following questions that relate to the chemistry of nitrogen.

(a) Two nitrogen atoms combine to form a nitrogen molecule, as represented by the following equation.

2 N(g) → N2(g)

Using the table of average bond energies below, determine the enthalpy change, ∆H, for the reaction. (2)

|Bond |Average Bond Energy (kJ mol–1) |

|N–N |160 |

|N=N |420 |

|N≡N |950 |

∆H = –950 kJ mol–1 ∆H = Σ(B.E.)reactants – Σ(B.E.)products

A triple bond is formed which is ΔH = [no bonds] − 1[N≡N]

an exothermic process. OR

ΔH = (0) − (950)

∆H = –950 kJ mol–1

(b) The reaction between nitrogen and hydrogen to form ammonia is represented below.

N2(g) + 3 H2(g) → 2 NH3(g) ∆ H˚ = –92.2 kJ/molrxn

Predict the sign of the standard entropy change, ∆S˚, for the reaction.

Justify your answer. (2)

∆S˚ is negative. There are fewer moles of product gas (2 mol) than reactant gas (4 mol) so the reaction is becoming more ordered (or less disordered) (or fewer microstates available).

(c) The value of ∆Go for the reaction represented in part (b) is negative at low temperatures but positive at high temperatures. Explain. (1)

ΔG° = ΔH° – TΔS°

(?) = (–) – T (–)

(?) = (–) (+)

At low T, the positive entropy term (–T∆S˚) is smaller than the negative enthalpy term (∆H˚), so ∆G˚ is negative.

At high T, the positive entropy term (–T∆S˚) is larger than the negative enthalpy term (∆H˚), so ∆G˚ is positive.

N2(g) + 2 H2(g) ↔ N2H4(g) ∆Ho298 = +95.4 kJ/molrxn ; ∆So298 = −176 J/molrxn·K

4. Answer the following questions about the reaction represented above using principles of thermodynamics.

(a) On the basis of the thermodynamic data given above, compare the sum of the bond strengths of the reactants to the sum of the bond strengths of the product. Justify your answer. (2)

Bond energy (B.E.) of reactants is greater than bond energy of products.

The reaction is endothermic, so more energy is absorbed to break bonds of reactants than is released when new bonds form in products according to the equation: ∆H = Σ(B.E.)reactants – Σ(B.E.)products

(b) Does the entropy change of the reaction favor the reactants or product?

Justify your answer. (2)

Entropy change favors reactants.

Since there are 3 moles of reactant gas compared to only 1 mole of product gas, there are more microstates of reactant molecules compared to product molecules (or more disorder in reactants than products).

(c) For the reaction under the conditions specified, which is favored, the reactants or the

product? Justify your answer. (2)

Reactants are favored because ∆G° for the reaction is positive.

ΔG° = ΔH° – TΔS°

(?) = (+) – T (–)

(+) = (+) (+)

A positive ∆H ° and a negative ∆S° yield a ∆G° that is always positive, independent of temperature.

(d) Explain how to determine the value of the equilibrium constant, Keq , for the reaction.

(Do not do any calculations.) (1)

Solve the equation ∆G° = −RT ln K for K.

Keq = e(–(G/RT)

Substitute the value of ∆G° (calculated from ΔG° = ΔH° – TΔS°), the value of T (298 K), and the value of R (8.314 J mol−1 K−1).

(e) Predict whether the value of Keq for the reaction is greater than 1, equal to 1,

or less than 1. Justify your answer. (1)

∆G° > 0, so K < 1. ∆G° = −RT ln K

OR ∆G° is positive so K must be less than 1 because ln K is negative which is multiplied by a negative for a positive ∆G° .

Answer KEY

1. 2001 2.

2. 2002 B 3.

3. 2003 7.

4. 2004 B 7.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download