Q54.For the reaction C2H6 →C2H4 + H2 the reaction enthalpy ΔrH in kJmol−1 is (Round off to the Nearest Integer). [Given : Bond enthalpies in kJmol−1 : C −C : 347, C = C : 611; C −H : 414, H −H : 436]
What This Question Tests
This question requires calculating the enthalpy of a reaction using given bond enthalpies, involving breaking and forming specific covalent bonds.
Concepts Tested
Formulas Used
ΔrH = Σ(Bond enthalpies of reactants) - Σ(Bond enthalpies of products)
📚 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
3.20 — For The Decomposition Of Azoisopropane To Hexane And Nitrogen At 543
Chemistry Class 11 · Chapter 3
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
- Enthalpy of reaction from bond enthalpies
- Year
- 2021
- Shift
- 18 Mar Shift 1
- Q Number
- Q54
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 6: Thermodynamics
More from this Chapter
Q89.A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2 . Work done by the engine is J (Q1 + Q2). This data (1) violates 1st law of thermodynamics (2) violates 1st law of thermodynamics if Q1 is −ve (3) violates 1st law of thermodynamics if Q2 is − ve(4) does not violate 1st law of thermodynamics
Q93.If at 298 K the bond energies of C −H, C −C, C = C and H −H bonds are respectively 414, 347, 615 and 435 kJ mol−1 , the value of enthalpy change for the reaction H2C = CH2( g) + H2( g) →H3C −CH3( g) at 298 K will be (1) −250 kJ (2) +125 kJ (3) −125 kJ (4) +250 kJ
Q94.In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria (1) (dS)V,E > 0, (dG)T,P < 0 (2) (dS)V:E = 0, (dG)TPP = 0 (3) (dS)V,E = 0, (dG)T,P > 0 (4) (dS)V,E < 0, (dG)T,P < 0
Q96.The enthalpy change for a reaction does not depend upon (1) use of different reactants for the same product (2) the nature of intermediate reaction steps (3) the differences in initial or final temperatures of (4) the physical states of reactants and products involved substances