4. For the following reaction Kp is 6.8 at 25°C and 1.21 x 10-3 at 200°C.2 NO2(g) → N2O4(g)a) What is the reaction quotient Q for the reaction? If Pn204 = 1 atm at 25°C what isPNO2?b) What is AG° at 25°C for this reaction?c) Determine AH at 25°C assuming it to be independent of temperature.d) Determine AS° at 25°C assuming it to be independent of temperature.

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Answered: 4. For the following reaction Kpis 6.8…

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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For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partial pressures (Kp). The value of Ke for the reaction shown below is equal to 3.2x104 at 298 K. 2NOBI (g) = 2NO(g)+Br,(g) M See Periodic Table O See Hint What is the value for Kp at 298 Kfor the reaction represented by the equation shown below? The value for the gas constant, R, is 0.08206 L• atm/mol • K. NOB:(g) = NO (g) +-Br, (g) (3)-1g - (3)ON
Determine the equilibrium constant for the system N2O4 ⇌ 2NO2 at 25°C. The concentrations are shown here: [N2O4] = 2.32 x 10^–2 M, [NO2] = 1.41 x 10^–2 M. a) 0.608 b) 1.65 c) 0.369 d) 8.57 x 10^-3 e) 1.17 x 10^2
For the generic reaction shown, K = 0.355 (at 950 K). What can you conclude about AG and AG° in a reaction mixture for which Q = 0.205 at 950 K? 2AZ2(g) + Z2(g) = 2AZ3(g) Item equation/constant: AG = AG° + RT In Q Select one: O a. AG is positive and AG° is negative O b. AG and AG° are both negative O c. AG is negative and AG° is positive O d. AG and AG° are both positive
Which equilibrium constant expression(s) are for the following reaction. (X stands for the mole fraction of the indicated substance in its phase, which we have so far assumed equal to unity (one) for essentially pure phases.) SnO2(s) + 2H2(g) ↔ Sn(s) + 2H2O(l) Choose from the list below. A) (XSnO2)eq I) (XSnO2)eq2 B) (pH2)eq J) (pH2)eq2 C) (XSn)eq K) (XSn)eq2 D) (XH2O)eq L) (XH2O)eq2 E) (XSnO2)eq-1 M) (XSnO2)eq-2 F) (pH2)eq-1 N) (pH2)eq-2 G) (XSn)eq-1 O) (XSn)eq-2 H) (XH2O)eq-1 P) (XH2O)eq-2
For the first one the options are increases, decreases, and remains the same.
Which causes the system below, at equilibrium, to shift to the product side? 2C09) +10 > 2C0) , AH° = -283.3 kJ O Removing carbon monoxide O Increasing temperature O Adding carbon dioxide ORemoving oxygen O Increasing pressure
> Given the two reactions HS (aq) +H (aq), K₁=9.70x108, and 1 1. H₂S(aq) 2. HS (aq) S² (aq) +H(aq), K₂=1.55x10-19, what is the equilibrium constant Kfinal for the following reaction? Enter your answer numerically. ▸View Available Hint(s) 1D ΑΣΦΑ Kfinal 1.078 1026 Submit ● Part R Previous Answers S² (aq) + 2H* (aq) = H₂S(aq) X Incorrect; Try Again; 9 attempts remaining ?
If K for the reverse reaction is 1.85 x 10² at 425°C, what is K for the forward C C reaction at the same temperature? -1.85 x 10² 54.1 1.85×10² 3.70 x 102 Request Answer Incorrect; Try Again; 5 attempts remaining Provide Feedback. Part A Given the reaction: 2HI=H₂ + I 2 ▼ Part A Given the reaction: 2HI H₂ + I2. If Ke for the reverse reaction is 1.85 x 102 at 425°C, what is Ke for the forward reaction at the same temperature? -1.85 x 10-2 54.1 1.85 x 102 O 3.70 x 10-² Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Provide Feedback
The equilibrium constant for a certain reaction decreases from 8.87 to 3.25 x 10-2 when the temperature increases from 25°C to 75°C. Hint: Manipulate the following equation. In(K) = -AH° ASO + ᎡᏆ R a Estimate the value of AS° for this reaction. AS⁰ = J/K - mol
17.12 Balance each reaction and write its reaction quotient, Q.: (a) NO(g) + O(g) = N,0;(g) (b) SF(g) + SO;(g) (c) SCIF,(g) + H2(8) = SO,F,(g) = S¿F10(8) + HCI(8) 0 bolb quotient 0:
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