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Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer?.

Solutions for Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.

Here you can find the meaning of Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer?, a detailed solution for Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? has been provided alongside types of Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Direction (Q. No. 11) This section contains 4multiple choice questions. Each question has fourchoices (a), (b), (c) and (d), out of whichONE or MORE THANT ONE is correct.Q.Variation o f equilibrium constant K with tem perature, T is given by van’t Hoff equationGraphically it is shown as given θ = tan (0 .5) and OP = 10Thus,a)pre-exponential factor is 10b)pre-exponential factor is 1010c)heat of reaction is 2853 J mol-1 at 298 Kd)heat of reaction is 9.574 J mol-1 at 298 KCorrect answer is option 'B,D'. Can you explain this answer? tests, examples and also practice Class 11 tests.