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2O3(g) ⇌ 3O2(g)
At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)
[Given: In 1.35 = 0.3 and R = 8.3 J K-1 mol-1]
At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)
[Given: In 1.35 = 0.3 and R = 8.3 J K-1 mol-1]
Correct answer is '747'. Can you explain this answer?
Most Upvoted Answer
2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard...
Given equation: 2O3(g) ⇌ 3O2(g)
At 300 K, ozone is fifty percent dissociated. This means that the equilibrium constant Kp = (PO2)^3 / (PO3)^2 = 1/2
Using the relationship ΔG° = -RTlnKp, we can calculate the standard free energy change at 300 K and 1 atm pressure.
ΔG° = -RTlnKp = - (8.3 J K^-1 mol^-1) x (300 K) x ln(1/2) = 747 J mol^-1 (nearest integer)
Therefore, the standard free energy change at 300 K and 1 atm pressure is -747 J mol^-1.
At 300 K, ozone is fifty percent dissociated. This means that the equilibrium constant Kp = (PO2)^3 / (PO3)^2 = 1/2
Using the relationship ΔG° = -RTlnKp, we can calculate the standard free energy change at 300 K and 1 atm pressure.
ΔG° = -RTlnKp = - (8.3 J K^-1 mol^-1) x (300 K) x ln(1/2) = 747 J mol^-1 (nearest integer)
Therefore, the standard free energy change at 300 K and 1 atm pressure is -747 J mol^-1.
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2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard...
Given, x = 0.5
= -2.303 × 8.3 × 300 log 1.35
= -8.3 × 300 ln 1.35
= -747 J mol-1
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2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer?
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2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer?.
2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer?.
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2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer?, a detailed solution for 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? has been provided alongside types of 2O3(g)3O2(g)At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-) ____ J mol-1. (Nearest integer)[Given: In 1.35 = 0.3 and R = 8.3 J K-1mol-1]Correct answer is '747'. Can you explain this answer? theory, EduRev gives you an
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