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The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer?.

Solutions for The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer?, a detailed solution for The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? has been provided alongside types of The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The surface of copper gets tarnished by the formation of copper oxide. N2gas was passed to prevent theoxide formation during heating of copper at 1250 K. However, the N2gas contains 1 mole % of watervapour as impurity. The water vapour oxidises copper as per the reaction given below:is the minimum partial pressure of H2(in bar) needed to prevent the oxidation at 1250 K. The value ofln is ____.(Given: total pressure = 1 bar, R (universal gas constant) = 8 J K-1mol-1, ln(10) = 2.3. Cu(s) and Cu2O(s)are mutually immiscible.At 1250 K: 2Cu(s) + ½ O2(g) -->Cu2O(s); ΔGθ= − 78,000 J mol−1H2(g) + ½ O2(g) --> H2O(g); ΔGθ= − 1,78,000 J mol−1; G is the Gibbs energy)Correct answer is '-14.6'. Can you explain this answer? tests, examples and also practice JEE tests.