5. (18 pts: 6 pts each) For a reaction:N2O4 (g) 2 NO2 (g), the thermodynamicconstants at 25°C are listed in the table below:CompoundNO2 (g)AH (kJ/mol)AG+ (kJ/mol)So (J/mol·K)33.2051.30240N2O4 (g)9.1697.82304.2(a) What is the equilibrium constant K of this reaction at 25°C?(b) In a closed container, the partial pressures of N2O4 (g) and NO2 (g) initially are 0.10 atmand 0.10 atm, respectively. Is this reaction spontaneous (i.e. moving forward) at 25°C?(c) Continued from (b), what will be the partial pressures of N2O4 and NO2 when the systemreaches equilibrium? (Help: Root of a quadratic equation is in Supporting Table-1.)

Given: T=298.15KΔHf°,kJ/molS°,J/molKN2O4(g)9.16304.2⇌2NO2(g)33.20240 A. Equilibrium…

Answered: 5. (18 pts: 6 pts each) For a reaction:…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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3.) Consider the reaction below. Reaction: AH°r (kJ/mole) 2 ICI (£) 17.78 L. (g) 247.551 l16.135 223.066 (/mole K) Which explains the outcome when 0.45 mol ½, 0.45 mol Cl2, and 2.42 mol ICl are added to a 2.0L flask at 25°C? A. [ICI] decreases B. [ICI] increases C. both [b] and [CL] increase D. at equilibrium Keywords: reaction quotient (Q), initial concentration(s), equilibrium concentration(s), equilibrium constant, forward reaction, reverse reaction, Q K. Q= K Concepts: Reaction Quotient, Chemical equilibrium
Calculate the Gibbs free energy change (∆GRXN) in kJ that occurs when one mole of solid iron (III) oxide is produced from solid iron and water. (One decimal place) The unbalanced chemical equation for the process is: Fe (s) + H2O (g) --> Fe2O3 (s) + H2 (g)
Compound SO; (g) SO₂ (g) NH3 (g) NO (g) H₂O (9) Consider the following reaction at 25 °C: 5 SO3(g) + 2 NH³(g) → 2 NO(g) + 5 SO₂(g) + 3 H₂O(g) Given the information in the table, calculate AH° for the reaction. AHf (kJ/mol) -396 -297 -46 90.4 -241.8 Sf (J/mol-K) 256 248 192.5 210.6 188.7
Methane-producing bacteria convert liquid acetic acid (CH3CO₂H) into CO₂(g) and CH₂(g). AS° ΔΗ, Ο (kJ/mol) (J/mol.K) CH3CO₂H (1) -484.5 159.8 CO₂(g) -393.5 213.8 CH₂(g) -74.8 186.2 Calculate AHOrxn = kJ Calculate AG rxp = (round to 3 sig figs) Is this process endothermic or exothermic under standard conditions? Is the reaction spontaneous under standard conditions? kJ (round to 3 sig figs)
please answer all the questions given in the picture.
Use the molar DHf° under the formulas to calculate DH°rxn for the equations as balanced: 1. 2B2H6(g) + 3CO2(g) --> 2B2O3(s) + 3CH4(g) ΔH°rxn = _______ kJ ΔH°f = +36 –394 –1274 –75 kJ/mol exo ? endo_thermic 2. 2P2O5 + 2CaC2 -->P4 + 2CaCO3 + 2CO2 ΔH°rxn = _______ kJ ΔH°f = –1505 –59 ___ –1207 –394 kJ/mol exo ? endo_thermic 3. 2Na2CrO4(s) + 10HCl(g) --> 4NaCl(s) + 3Cl2(g) + Cr2O3(s) + 5H2O(l) ΔH°f = –1342 –92 –411 ___ –1140 –286 kJ/mol ΔH°rxn = _______ kJ exo ? endo_thermic
Calculate the Gibbs free energy change (∆GRXN) in kJ that occurs when one mole of solid iron (III) oxide is produced from solid iron and water. (One decimal place) The unbalanced chemical equation for the process is: Fe (s) + H2O (g) --> Fe2O3 (s) + H2 (g)
Methane-producing bacteria convert liquid acetic acid (CH³CO₂H) into CO₂(g) and CH₂(g). AS° ΔΗ, Ο (kJ/mol) (J/mol.K) CH3CO₂H (1) -484.5 159.8 CO₂(g) -393.5 213.8 CH₂(g) -74.8 186.2 Calculate AHºrxn = kJ Calculate AG, = rxn (round to 3 sig figs) Is this process endothermic or exothermic under standard conditions? Is the reaction spontaneous under standard conditions? kJ (round to 3 sig figs)
For the reaction C(s,graphite) + 02(g) CO2(g) AG° = -394.3 kJ and AS° = 2.9 J/K at 290 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 290 K. The standard enthalpy change for the reaction of 2.38 moles of C(s,graphite) at this temperature would be kJ.
Heat of reaction (AH°) for the following reaction at 25.0 °C is Fe;O4(6) + CO(g) 3FEO(s) + CO2(g) Given the following Heat of Formation: AHF0304(s) = (-1118 kJ/mol), AHCO(g) = (-110.5 kJ/mol), AHF«06) = (-272KJ/mol), AHco2(g) ( -393.5 kJ/mol),

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