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Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer?.

Solutions for Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.

Here you can find the meaning of Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Rate constant k of a reaction varies, with temperature according to the equation log k = constant – Ea/2.303R x 1/T where Eais the energy of activation for the reaction When a graph is plotted for log k vs 1/T a straight line with a slope -6670 k is obtained. The activation energy for this reaction will be (R = 8.314JK-1mol-1)a)122.65 kJmol-1b)127.71 kJmol-1c)142.34 kJmol-1d)150.00 kJmol-1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice NEET tests.