2) Calculate the numeric value of the equilibrium constant K for reaction at 298 K:2 NO2 (g) → N₂O4 (g)from Data Section:les of das Athina.pdf-Adebe Acrobat Reader (54-b0)You SignWindow HToolsHome827-31.40 -Tables of data Atkin xaTable 2.7 (Continued)NeonNe(g)NitrogenN₂(g)N(g)NO(g)N₂O(g)NO₂(g)N₂O₂(g)N₂O5(s)N₂O,(g)1081 (12 of 4)TTATO (11AM/(g mol-¹)20.1828.01314.00730.0144.0146.0192.1108.01108.01OO2018. PAH*/(kJ mol-¹)00+472.70+90.25+82.05+33.18+9.16-43.1+11.3174 17AG/(kJ mol-¹)00+455.56+86.55+104.20+51.31+97.89+113.9+115.1℗30 31xSign In&00SBD2C%

Given : ∆G°f for NO2 = +51.31 ∆G°f for N2O4 = +97.89 The balance chemical reaction is 2 NO2 (g)…

2) Calculate the numeric value of the equilibrium constant K for reaction at 298 K: 2 NO2 (g) → N₂O4 (g) from Data Section: of inspf-Adobe Acrobat Reader - The Spider 2 Tables of data Arkin, Table 2.7 (Continued) Neon Ne(g) Nitrogen N₂(g) N(g) 1001 (24) NO(g) N₂O(g) NO₂(g) N₂O4(g) N₂O5(s) N₂O(g) TRATO (11 M/(g mol-¹) 20.18 28.013 14.007 30.01 44.01 46.01 92.1 108.01 108.01 PRADEG AH*/(kJ mol-¹) 0 +472.70 +90.25 +82.05 +33.18 +9.16 -43.1 +11.3 AG/(kJ mol-¹) 0 0 +455.56 +86.55 +104.20 +51.31 +97.89 +113.9 +115.1 00.31 e a

2) Calculate the numeric value of the equilibrium constant K for reaction at 298 K: 2 NO2 (g) → N₂O4 (g) from Data Section: of inspf-Adobe Acrobat Reader - The Spider 2 Tables of data Arkin, Table 2.7 (Continued) Neon Ne(g) Nitrogen N₂(g) N(g) 1001 (24) NO(g) N₂O(g) NO₂(g) N₂O4(g) N₂O5(s) N₂O(g) TRATO (11 M/(g mol-¹) 20.18 28.013 14.007 30.01 44.01 46.01 92.1 108.01 108.01 PRADEG AH*/(kJ mol-¹) 0 +472.70 +90.25 +82.05 +33.18 +9.16 -43.1 +11.3 AG/(kJ mol-¹) 0 0 +455.56 +86.55 +104.20 +51.31 +97.89 +113.9 +115.1 00.31 e a

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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10:26 1 4 7 +/- For the reaction 2 A (g) → B (g), Kp = 0.00617 at 298 K. When AG = 8.35 kJ/mol, what is the partial pressure of B when the partial pressure of A is 2.00 atm for this reaction at 298 K. 2 5 Question 12 of 15 8 atm 3 6 9 0 al 9% Tap here or pull up for additional resources ||| O Submit Xx C x 100
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6. Consider the following reaction: CO(g) + H2O(g) CO2(g) + H2(g) Ko = 0.58 at 1273 K* A sealed 4.00L vessel initially contains 1.00 mole CO2(g), 1.00 mole H2(g), 0.600 moles CO(g), and 0.600 moles of H2O(g). Calculate the moles of H2(g) present at equilibrium. The NG Reference: General Chemistry 8th Ed: Ebbing and Gammon: 2005 by Houghton Mifflin. New York. page 654, print 68
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q The equilibrium constant, Ke, for the following reaction is 1.80 x 104 at 298 K. [Review Topics] [References] Use the References to access important values if needed for NH4HS(9) NH3(g) + H₂S(9) Calculate the equilibrium concentration of H₂S when 0.322 moles of NH4HS(s) are introduced into a 1.00 L vessel at 298 K. [H₂S] = M Submit Answer Show Hint NOV 30 Le... The equilib tv Retry Entire Group 9 more group attempts remaining Email In M
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Calculate the value of K, for the reaction 2N₂(g) + O₂(g) 2N₂O(g) at 298.15 K and 1473 K. Thermodynamic data for N₂O(g) are: AH = 82.05 kJ/mol; Sº = 219.9 J/mol -K; AG% = 104.2 kJ/mol. 1st attempt Part 1 2 298.15 K Part 2 1473 K