Q53.The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are: -2220 . 0 kJ mol-1, -393 . 5 kJ mol-1 and -285 . 8 kJ mol-1 respectively. The magnitude enthalpy of formation of propane C3H8 is _____ kJmol-1. (Nearest integer)
What This Question Tests
This question uses Hess's Law to calculate the enthalpy of formation of propane from its elements, given the standard enthalpies of combustion of propane, graphite, and dihydrogen.
Concepts Tested
Formulas Used
ΔrH = ΣΔfH(products) - ΣΔfH(reactants)
ΔfH = ΣΔcH(reactants) - ΣΔcH(products) (modified for specific reaction)
📚 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
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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.
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3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.
📋 Question Details
- Chapter
- Thermodynamics & Thermochemistry
- Topic
- Hess's Law
- Year
- 2022
- Shift
- 25 Jul Shift 1
- Q Number
- Q53
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 6: Thermodynamics
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