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For a given reaction, ΔH = 35.5 kJ mol-1 and ΔS = 83.6 JK-1 mol-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)a)T > 425 Kb)All temperaturesc)T > 298 Kd)T < 425 KCorrect answer is option 'A'. 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 For a given reaction, ΔH = 35.5 kJ mol-1 and ΔS = 83.6 JK-1 mol-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)a)T > 425 Kb)All temperaturesc)T > 298 Kd)T < 425 KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a given reaction, ΔH = 35.5 kJ mol-1 and ΔS = 83.6 JK-1 mol-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)a)T > 425 Kb)All temperaturesc)T > 298 Kd)T < 425 KCorrect answer is option 'A'. Can you explain this answer?.

Solutions for For a given reaction, ΔH = 35.5 kJ mol-1 and ΔS = 83.6 JK-1 mol-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)a)T > 425 Kb)All temperaturesc)T > 298 Kd)T < 425 KCorrect answer is option 'A'. 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.

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