(a)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy is represented by
If the
Answer to Problem 6.38P
The formation of starting material is favored at the given value of
Explanation of Solution
Given
The value of
The given value of
(a) The formation of starting material is favored at the given value of
(b)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by
The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.
Answer to Problem 6.38P
The formation of the productis favored at the given value of
Explanation of Solution
Given:
The value of
The given value of
The formation of the product is favored at the given value of
(c)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by
The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.
Answer to Problem 6.38P
The formation of the starting material is favoredat the given values of
Explanation of Solution
Given
The values of
The given values of
The formation of the starting material is favouredat the given values of
(d)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy is represented by
If the
Answer to Problem 6.38P
The formation of the productis favored at the given value of
Explanation of Solution
Given
The value of
The given value of
The formation of the product is favored at the given value of
(e)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by
The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.
Answer to Problem 6.38P
The formation of the starting materialis favored at the given value of
Explanation of Solution
Given
The value of
The given value of
The formation of the starting material is favored at the given value of
(f)
Interpretation: The starting material or product which is favored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by
The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.
Answer to Problem 6.38P
The formation of the productis favored at the given value of
Explanation of Solution
Given
The value of
For the spontaneous reaction, the value of
The formation of the product is favored at the given value of
(g)
Interpretation: The starting material or product which isfavored at equilibrium is to be identified.
Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by
The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.
Answer to Problem 6.38P
The formation of the starting material is favored at the given value of
Explanation of Solution
Given
The value of
The value of
The formation of the starting material is favored at the given value of
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Chapter 6 Solutions
ORG.CHEMISTRY CONNECT ACCESS>CUSTOM<
- What is the ΔH of the reaction 2KS+2K→2K2S?K+S→KS ΔH=+32.5 kJKS+K→K2S ΔH=+38.2 kJ A +70.7 kJ B +65 kJ C +6 kJ D +76.4 kJarrow_forward1. The reaction _______(showin in a/b)______ is endothermic/exothermic and therefore heat is absorbed/released by the reaction. a) PCl5 -> PCl3 + Cl2 delta H = +67.0 kJ b) PCl3 + Cl2 -> PCl5 delta H = -87.0 kJ state if a and b are endothermic or exothermic and absorbed or released and why.arrow_forwardCalculate the ∆G in kJ/mol for a reaction at room temperature (~25.00°C) that has a Keq = 1000.0. Use R = 8.3145 J/mol·K.arrow_forward
- The reaction CBr4(g) + H2(g) → CHBr3(g) + HBr(g) occurs at room temperature (20oC) with ΔHo = -137.8 kJ and ΔSo = −55.0 J/K. What is ΔSuniverse (in J/K.mol) for this reaction?arrow_forwardWhich reaction had a negative delta T value l? Is the reaction endothermic or exothermic explainarrow_forwardFor which reaction is ΔG° expected to be closest to ΔH°?N2(g) + 3H2(g) ⇄ 2NH3(g)H2O(ℓ) ⇄ H2O(s)NaCl(s) ⇄ Na+(aq) + Cl-(aq)CO2(g) ⇄ CO2(s)2NO(g) ⇄ N2(g) + O2(g)arrow_forward
- Calculate So values for the following reactions by using tabulated So values from Appendix C.(a) 2 NOCl(g) Æ 2 NO(g) + Cl2(g)(b) Be(OH)2(s) BeO(s) + H2O(g)(c) HCl(g) + NaOH(aq) NaCl(s) + H2O(l)(d) 2 CH4(g) C2H6(g) + H2(g)arrow_forwardCalculate ∆S° for the following reactions in J K-1 a. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)b. Ca(OH)2(s) + H2SO4(l) → CaSO4(s) + 2 H2O(l)arrow_forwardConsidering each of the following values and neglecting entropy, tell whether the starting material or product is favored at equilibrium: (a) ?H° = 80 kJ/mol; (b) ?H° = -40 kJ/mol.arrow_forward
- Calculate the nonstandard free energy change, ΔG, for the following reaction when P I2 = 2.50 atm, P Cl2 = 3.25 atm and P ICl = 0.210 atm. I2(g)+Cl2(g)⇌ICl(g) ΔG∘=−10.9 kJ/molarrow_forwardIf Kp = 947 for the reaction below at 310.0 K, then what is the value of Kc?(R = 0.0821 L・atm/mol・K.) 2 A (g) + B (s) ⇌ 2 C (s) + D (g)arrow_forwardGiven each value, determine whether the starting material or product is favored at equilibrium. a. Keq = 0.5 b. ΔG° = −100 kJ/mol c. ΔH° = 8.0 kJ/mol d. Keq = 16 e. ΔG° = 2.0 kJ/mol f. ΔH° = 200 kJ/mol g. ΔS° = 8 J/(K · mol) h. ΔS° = −8 J/(K · mol)arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning