Start Free Mock LoginAlready registered? Login here

Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionHard2 minPYQ_2007

ChemistryHardsingle choice

Given that bond energies of $H - H$ and $Cl - Cl$ are $430 kJ mol^{- 1}$ and $240 kJ$ $mol^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- kJ$ $mol^{- 1}$ , bond enthalpy of $HCl$ is:NEET 2007

Options:

Answer:

Solution:

$\frac{1}{2} H_{2} + \frac{1}{2} Cl_{2} ⟶ HCl$ $Δ H_{HCl} = Σ$ B.E. of reactant $- Σ$ B.E. of products $- 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 -$ B.E. of $HCl$ $∴$ B.E. of $HCl = 215 + 120 + 90$ $= 425 kJ mol^{- 1}$

Stream:NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2007

Doubts & Discussion

Loading discussions...