HOMEWORK CHAPTER 18
COSUMNES RIVER COLLEGE
Direction: For exercises that require problem solving, you need to show the calculation set up leading to the answer in order to receive credit. Make sure your answers contain the correct units. YOUR WILL NEED TO USE A TABLE OF THERMODYNAMIC VALUES TO SOLVE SOME PROBLEMS.
- Which substance has the higher entropy ? Provide an explanation for credit.
- Dry ice (solid CO2) at –78 C or CO2(g) at 0 C.
- One mole of N2(g) at 1 bar pressure or one mole of N2(g) at 10 bar pressure (both at 25 C).
- A sample of pure silicon (to be used in a computer chip) or a piece of silicon containing a trace of another element such as boron or phosphorus.
- methane gas, CH4(g), or carbon tetrachloride gas, CCl4(g)
- Sulfur dioxide gas, SO2(g), or butane gas, CH3CH2CH2CH3(g)
- Use S values to calculate the standard entropy change, S, for each of the following changes and comment on the sign of S (i.e. why is it positive or negative ?).
- CCl4(g) → CCl4()
Answer: S = −93.3 J/molK
(Question #2 continued)
- NH4Cl(s) → NH4Cl(aq)
Answer: S = 73.2 J/molK
- Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
Answer: Srxn = 266.9 J/molK
- Is the reaction Si(s) + 2 Cl2(g) → SiCl4(g) spontaneous under standard conditions at 298 K ? Answer this question by calculating Ssys, Ssurr, and Suni. [Here, the system is the reaction and includes reactants and products; SiCl4(g): S = 330.9 J/molK, Hf = −662.8 kJ/mol]
Answer: Suni = 2090.1 J/molK
- For each of the following reactions, determine which one is 1) spontaneous at all temperatures, 2) nonspontaneous at all temperatures, 3) more favorable at lower temperatures, or 4) more favorable at higher temperatures. Provide an explanation for each case for credit.
- N2(g) + 2 O2(g) → 2 NO2(g) Hrxn = 66.2 kJ/mol Srxn = – 121.6 J/molK
(Question #4 continued)
- C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O() H rxn = – 673 kJ/mol S rxn = 60.4 J/molK
- Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s) Hrxn = – 851.5 kJ/mol Srxn = – 375.2 J/molK
- MgO(s) + C(graphite) → Mg(s) + CO(g) Hrxn = 490.7 kJ/mol Srxn = 197.9 J/molK
- Using values of Hf and S, first calculate Hrxn and Srxn, and use these to calculate Grxn for each of the following reactions. Is the reaction spontaneous under standard conditions ? Is it product-favored or reactant- favored at equilibrium ? Is it entropy-driven or enthalpy-driven ?
- 2 Na(s) + 2 H2O() → 2 NaOH(aq) + H2(g)
Answer: Grxn = −363.9 kJ/mol
(Question #5 continued)
- H2O(l) → H2(g) + ½ O2(g)
Answer: Grxn = 237.1 kJ/mol
- SnO2(s) + C(s, graphite) → Sn(s, white) + CO2(g)
Answer: Grxn = 121.4 kJ/mol
Is the above reaction spontaneous under standard conditions ? If not, at what temperature, in C, does it become spontaneous ?
Answer: T = 602 C
- For the reaction: TiCl2(s) + Cl2(g) → TiCl4() Grxn = – 272.8 kJ/mol TiCl2
Using this information and other data available in the Table of Thermodynamic Values, calculate the value of Gf, in kJ/mol, for TiCl2(s).
Answer: GfTiCl2(s) = −464.4 kJ/mol
- Consider the following reaction: 3 O2(g) 2 O3(g) Grxn = 163.2 kJ/mol
Calculate Kp for the reaction at standard temperature (25 C). Comment on the sign of Grxn and the magnitude of Kp (Is the reaction spontaneous as written ? Is it reactant-favored or product-favored ?)
Answer: Kp = 2.5 x 10−29
- The formation of 1 mole of Fe2O3(s) from its elements is depicted in the following balanced equation:
2 Fe(s) + 3 O2(g) → Fe2O3(s)
- What is the standard free energy of formation (Gf) of 1.00 mole of Fe2O3(s) ? (Look up this value in the Table of Thermodynamic Values.
- What is the value of G when 454 g of Fe2O3(s) is formed from the elements ?
Answer: G = −2110 kJ
- Calculate the entropy change, S, for the vaporization of ethanol, C2H5OH, at its normal boiling point, 78.0 C.
C2H5OH(l) C2H5OH(g) Hvap = 39.3 kJ/mol
Note: During phase transition, such as liquid boiling to gas or solid melting to liquid and vice versa, the two phases are in equilibrium with one another. Under these equilibrium conditions, G = 0
Answer: S = 112 J/molK
- Some metal oxides can be decomposed to the metal and oxygen under reasonable conditions. An example is the decomposition of silver oxide, Ag2O, to silver and oxygen:
2 Ag2O(s) → 4 Ag(s) + O2(g)
- Is the decomposition of silver oxide product-favored at 25 C (i.e. standard temperature) ? [There are more than one method to determine whether a reaction is product-favored. One way is to calculate the Grxn for the reaction under standard conditions. Look up appropriate values in the Table of Thermodynamic Values.]
Answer: Grxn = 22.4 kJ/mol
- If the reaction is nonspontaneous under standard conditions, can it become spontaneous if the temperature is raised ? At what temperature does the reaction become product-favored ?
Answer: At T = 195 C and above, the decomposition of Ag2O becomes spontaneous
- Indicate whether each of the following statements is CORRECT or INCORRECT, then explain your choice for credit.
- Entropy increases in all spontaneous reactions.
- All spontaneous processes are exothermic.
- Reactions with a negative free energy change (G < 0) are product-favored and occur with a rapid rate of transformation of reactants to products.
- Reactions with a positive H and a positive S can never be product-favored.
- If G for a reaction is negative, the reaction will have an equilibrium constant greater than 1.
- For each of the following processes, predict the algebraic signs of H, S, and G, and provide a brief explanation in each case for your choice. No calculations are necessary.
- The decomposition of liquid water to give gaseous oxygen and hydrogen, a process that requires a considerable amount of energy.
H2O(l) → ½ O2(g) + H2(g)
(Question #12 continued)
- The combustion of gasoline in the automobile engine is exemplified by the combustion of octane (C8H18):
2 C8H18() + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
- When solid ammonium nitrate (NH4NO3) dissolves in water, the water temperature gets colder.
- Calculate the free energy of formation (Gf, in kJ/mol) for HI(g) at 350 C: ½ H2(g) + ½ I2(g) HI(g)
given the following equilibrium partial pressures: PH2 = 0.132 bar
PI2 = 0.295 bar
PHI = 1.61 bar
Answer: Gf = − 10.9 kJ/mol
- Estimate the vapor pressure of ethanol, C2H5OH, at 37 C using the data below. Express the result in mmHg.
C2H5OH(l) C2H5OH(g) Hint: Determine Grxn for the reaction. Then
Gf(kJ/mol) −174.8 −167.9 compute Kp, which leads to PC2H5OH(g)
Answer: PC2H5OH(g) = 52.1 mmHg
- When vapors from aqueous hydrochloric acid and aqueous ammonia come in contact, they react to produce a white
“cloud” of solid NH4Cl:
HCl(g) + NH3(g) NH4Cl(s)
- Predict whether the sign for each of the following properties is less than 0, larger than 0, or equal 0, and explain your prediction for credit.
- Using values of Hf and S, calculate the values of Ssys, Ssurr, Suni, Gsys, and Kp for this reaction at
Answer: Ssys = −285.1 J/molK; Ssurr = 591.3 J/molK; Suni = 306.2 J/molK Gsys = −91.2 kJ/mol; Kp = 9.9 x 1015 (or rounded off to 1.0 x 1016)
- Living organisms use food energy to create energy-rich adenosine triphosphate (ATP), which then acts as an energy source for a variety of reactions that an organism must carry out to survive. ATP provides energy through its hydrolysis:
ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) Grxn = −30.5 kJ/mol ATP hydrolyzed
where ADP represents adenosine diphosphate and Pi an inorganic phosphate group (such as HPO42−).
- The food energy can be derived from the oxidation of glucose (C6H12O6) to form CO2 and H2O:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Using values of Gf, calculate the Grxn for the oxidation of 1 mole of glucose.
Answer: Grxn = −2878.6 kJ/mol glucose
- The free energy obtained glucose oxidation can be used to drive the formation of ATP from ADP by reversing the ATP hydrolysis reaction. Estimate the number of moles of ATP that can be formed from the oxidation of 1 mole of glucose.
Answer: 94.4 mol ATP
- Which process is necessarily driven by an increase in the entropy of the surroundings (Ssurr > 0) ? Explain for credit. [Hint: It may be helpful to first evaluate the entropy change of the system (Ssys); the systems are the processes shown below.]
- the condensation of water
- the sublimation of dry ice
(Question #17 continued)
- ice cube melts when left out at room temperature
- a gas enclosed in a piston-cylinder apparatus is being compressed at constant temperature by an external pressure
- For the reaction: A(g) → B(g) G = −42.5 kJ/mol (standard free energy change)
Under certain nonstandard conditions, the free energy change (G) for this reaction is zero. Which of the following statements (a → c) must be true about this reaction ? and which is false ? Show your work in the space below.
- The concentration of the product is greater than the concentration of the reactant.
- The reaction is at equilibrium
- The concentration of the reactant is greater than the concentration of the product.