Question Description
Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? 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 Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer?.

Solutions for Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.

Here you can find the meaning of Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the formation of CaO(s). Assume thatare independent of temperature.Calculatefor the fornation of CaO(s) at 0°C and is the reaction spontaneous or non-spontaneousat 0°C.a)-624.63 kJ mol-1, spontaneousb)624.63 kJ mol-1, non-spontaneousc)-646.37 kJ mol-1, spontaneousd)646.37 kJ mol-1, non-spontaneousCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice IIT JAM tests.