(a)
Interpretation:
The standard free-energy change at
Concept introduction:
The relation between change in standard free energy of ionization
Where,
•
•
•
Answer to Problem 3.49AP
The standard free-energy change at
Explanation of Solution
The standard free-energy change is given by the formula shown below.
Where,
•
•
•
The calculated value of
The temperature is given as
Substitute the temperature
Substitute
Therefore, the standard free-energy change at
The standard free-energy change at
(b)
Interpretation:
The standard free-energy change at
Concept introduction:
The relation between change in standard free energy of ionization
Where,
•
•
•
Answer to Problem 3.49AP
The standard free-energy change at
Explanation of Solution
The standard free-energy change is given by the formula shown below.
Where,
•
•
•
The calculated value of
The temperature is given as
Substitute the temperature
Substitute
Therefore, the standard free-energy change at
The standard free-energy change at
(c)
Interpretation:
The concentration of each species at equilibrium for
Concept introduction:
Equilibrium constant of a reaction is expressed by the ratio of concentration of product species raised to the power of their
Answer to Problem 3.49AP
The concentration of
Explanation of Solution
The reaction (a) in Problem 3.48 is given as shown below.
In the above reaction,
The initial concentration of
Let the concentration consumed by each reactant be
The equilibrium constant for the reversible reaction is written as shown below.
The calculated value of
Substitute the concentration of each species at equilibrium in equation (1).
Using the quadratic formula, the value of
Substitute the value of
Therefore, the two possible values of
Therefore, the value of
The concentration of
The concentration of
The concentration of
The concentration of
Therefore, the concentration of
The concentration of
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Chapter 3 Solutions
Organic Chemistry Study Guide and Solutions
- Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forwardUsing values of fH and S, calculate rG for each of the following reactions at 25 C. (a) 2 Na(s) + 2 H2O() 2 NaOH(aq) + H2(g) (b) 6 C(graphite) + 3 H2(g) C6H6() Which of these reactions is (are) predicted to be product-favored at equilibrium? Are the reactions enthalpy- or entropy-driven?arrow_forwarda Calculate K1, at 25C for phosphoric acid: H3PO4(aq)H+(aq)+H2PO4(aq) b Which thermodynamic factor is the most significant in accounting for the fact that phosphoric acid is a weak acid? Why ?arrow_forward
- Assuming that neither standard enthalpy changes of formations (∆Hof) nor standard molar entropies (So) depend upon temperature, estimate using the Table of Thermodynamic Data : (a) the standard Gibbs free energy change for the reaction that forms rhombic sulfur at 600 K, and (b) the temperature (in oC) at which reaction will stop formation of products: 2H2S(g) + SO2(g) → 3S(rhombic, s) + 2H2O(g) Round off your answers to the nearest integer. Report the temperature in oC. and enter them with correct units: (a)∆Gorxn = (b) T =arrow_forward(a) Calculate the standard free-energy change and the equilibrium constant for the dimerization of NO, to N2O4 at 25°C (see Appendix D). (b) Calculate AG for this reaction at 25°C when the pres- sures of NO2 and N,O4 are each held at 0.010 atm. Which way will the reaction tend to proceed?arrow_forwarda) As stated in question 5a, glucose (C6H12O6(s)) is a source of cellular energy. Calculate the standard free energy for the metabolism of glucose: Given ΔG°C6H12O6(s) = -911kJ/mol; ΔG°O2(g) = 0kJ/mol; ΔG°CO2 = -394kJ/mol; ΔG°H2O(l) = -237kJ/molC6H12O6(s) + 6O2(g) ➝ 6CO2(g) + 6H2O(l) b) Cells couple the hydrolysis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) andinorganic phosphate (PO43–) to drive chemical reactions (i.e. as a source of chemical energy). The reaction is: ATP(aq) + H2O(l) ➝ ADP(aq) + PO43–(aq) Calculate K for this reaction if ΔG° = –30.5 kJ/mole. c) If all of the energy from glucose metabolism went into ATP synthesis from ADP and inorganicphosphate, how many molecules of ATP could be generated from each molecule of glucose?arrow_forward
- A student determines the value of the equilibrium constant to be 3.67×10-18 for the following reaction. Fe3O4(s) + 4H₂(g) →→→3Fe(s) + 4H₂O(g) Based on this value of Keq: AGO for this reaction is expected to be (greater, less) than zero. Calculate the free energy change for the reaction of 1.67 moles of Fe3O4(s) at standard conditions at 298K. AGᵒrxn KJarrow_forward11. What is the ΔHº-value for the dissolution of NaOH (s) to yield 1.00... m NaOH (aq), given: Na (s) + 1/2 O2 (g) + 1/2 H2 (g) → NaOH (s) ΔHfº = -425.9 kJ/mol Na (s) + 1/2 O2 (g) + 1/2 H2 (g) → NaOH (aq, 1 m) ΔHfº = -469.2 kJ/mol Multiple Choice -895.1 kJ/mol -43.3 kJ/mol +1.102 kJ/mol +43.3 kJ/mol +895.1 kJ/molarrow_forwardConsider the following reaction: CH3OH(g) --> HCHO(g) + H2(g); DHo = 85 kJ (2) Which of the following statements is TRUE for this reaction when carried out under standard condition? (A) This reaction is spontaneous at all temperatures; (B) This reaction is not spontaneous at any temperature. (C) This reaction is spontaneous at high temperatures, but not at low temperatures; (D) This reaction is spontaneous at low temperatures, but not at high temperatures;arrow_forward
- Using the provided data table, calculate ΔrxnH for the following reactions:a) 2O3(g) ⇌ 3O2(g)b) H2S(g) + 3/2 O2(g) ⇌ H2O(ℓ) + SO2(g) Compound ΔfH (kJ mol–1)H2O(ℓ) –285.83H2S (g) –20.6O2 (g) 0O3 (g) 142.7SO2 (g) –296.81arrow_forward Determine ΔG°rxn for the following reaction: N2O (g) + NO2 (g) à 3 NO (g)ΔG° = ? Use the following reaction with known ΔG° values. Rxn 1:2 NO (g) + O2 (g) à 2 NO2 (g)ΔG1° = - 71.2 kJ Rxn 2:N2 (g) + O2 (g) à 2 NO (g)ΔG2° = + 175.2kJ Rxn 3:2 N2O (g) à 2 N2 (g) + O2 (g) ΔG3° = - 207.4kJarrow_forward3. (a) Use the data given below and calculate AHO, ASO, A Gº, and Kp at 25° C for the reaction: CO (g) + 3 H₂ (g) - → CH4 (g) + H₂O (g) (b) Calculate AG for the reaction at 250 °C. (c) At what temperature (°C) is AG equal to zero? In what temperature range is this reaction product- favored? Compound CO (g) H₂(g) CH4 (g) H₂O (g) AHO, kJ/mol -110.52 0 -74.81 -241.82 So, J/mol K 197.67 130.68 186.264 188.83arrow_forward
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