Q96.The enthalpies of combustion of carbon and carbon monoxide are −393.5 and −283 kJ mol−1 respectively. The enthalpy of formation of carbon monoxide per mole is (1) 110.5 kJ (2) −110.5 kJ (3) −676.5 kJ (4) 676.5 kJ
What This Question Tests
This question tests the ability to apply Hess's Law to calculate the enthalpy of formation of carbon monoxide using given enthalpies of combustion.
Concepts Tested
Formulas Used
ΔH_reaction = ΣΔH_products - ΣΔH_reactants
📚 NCERT Sections This Tests
3.23 — The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 × 10–5S–1
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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.
1.35 — Henry’S Law Constant For The Molality Of Methane In Benzene At 298 K Is
Chemistry Class 11 · Chapter 1
1.35 Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
3.26 — The Decomposition Of Hydrocarbon Follows The Equation
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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
📋 Question Details
- Chapter
- Thermodynamics & Thermochemistry
- Topic
- Hess's Law
- Year
- 2004
- Shift
- Unknown
- Q Number
- Q96
- Type
- MCQ
- NCERT Ref
- Class 11 Chemistry Ch 6: Thermodynamics
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