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For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is?
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For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information ...
Calculation of ∆Suniverse for the reaction H2(g) 1/2O2(g) → H2O(l)
Step 1: Calculate ∆Ssystem
∆Ssystem = ΣS(products) - ΣS(reactants)
∆Ssystem = S(H2O(l)) - [S(H2(g)) + 1/2S(O2(g))]
∆Ssystem = [70 J/K mol] - [(130 J/K mol) + (204 J/K mol)]
∆Ssystem = -264 J/K mol
Step 2: Calculate ∆Ssurroundings
∆Ssurroundings = -∆H/T
∆Ssurroundings = -(-285 kJ/mol)/(300 K)
∆Ssurroundings = 0.95 kJ/K mol
Step 3: Calculate ∆Suniverse
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse = (-264 J/K mol) + (0.95 kJ/K mol)
∆Suniverse = 686 J/K mol
Explanation
The entropy change of the system (∆Ssystem) is calculated using the standard molar entropies of the products and reactants. The entropy change of the surroundings (∆Ssurroundings) is calculated using the standard enthalpy change of the reaction (∆H) and temperature (T). The entropy change of the universe (∆Suniverse) is the sum of the entropy change of the system and surroundings. In this case, the positive value of ∆Suniverse indicates that the reaction is spontaneous in the forward direction at the given temperature.
Step 1: Calculate ∆Ssystem
∆Ssystem = ΣS(products) - ΣS(reactants)
∆Ssystem = S(H2O(l)) - [S(H2(g)) + 1/2S(O2(g))]
∆Ssystem = [70 J/K mol] - [(130 J/K mol) + (204 J/K mol)]
∆Ssystem = -264 J/K mol
Step 2: Calculate ∆Ssurroundings
∆Ssurroundings = -∆H/T
∆Ssurroundings = -(-285 kJ/mol)/(300 K)
∆Ssurroundings = 0.95 kJ/K mol
Step 3: Calculate ∆Suniverse
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse = (-264 J/K mol) + (0.95 kJ/K mol)
∆Suniverse = 686 J/K mol
Explanation
The entropy change of the system (∆Ssystem) is calculated using the standard molar entropies of the products and reactants. The entropy change of the surroundings (∆Ssurroundings) is calculated using the standard enthalpy change of the reaction (∆H) and temperature (T). The entropy change of the universe (∆Suniverse) is the sum of the entropy change of the system and surroundings. In this case, the positive value of ∆Suniverse indicates that the reaction is spontaneous in the forward direction at the given temperature.
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For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information ...
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For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is?
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For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is?.
For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction H2(g) 1/2O2(g) --> H2O(l). the following information is given T=300K ∆H=-285 KJ/mol SO2(g)=204 J/K mol SH2O(l)=70 J/K mol SH2(l)=130 J/K mol ∆Suniverse for the reaction is?.
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