Q59.If the activation energy of a reaction is 80. 9 kJ mol−1 , the fraction of molecules at 700 K, having enough energy to react to form products is e−x . The value of x is (Rounded off to the nearest integer) [Use R = 8. 31 J K−1 mol−1 ]
What This Question Tests
This question applies the Nernst equation to calculate the EMF of a given electrochemical cell at a non-standard concentration, requiring calculation of standard cell potential and reaction quotient.
Concepts Tested
Formulas Used
Ecell = E°cell - (0.059/n) logQ
E°cell = E°cathode - E°anode
📚 NCERT Sections This Tests
3.23 — The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 × 10–5S–1
Chemistry Class 11 · Chapter 3
3.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
3.26 — The Decomposition Of Hydrocarbon Follows The Equation
Chemistry Class 11 · Chapter 3
3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26
2.9 — The Resistance Of A Conductivity Cell Containing 0.001M Kcl Solution At 298
Chemistry Class 11 · Chapter 2
2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 59 Electrochemistry Reprint 2025-26
📋 Question Details
- Chapter
- Electrochemistry
- Topic
- Nernst equation
- Year
- 2021
- Shift
- 26 Feb Shift 2
- Q Number
- Q59
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 3: Electrochemistry
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