Ncert Exercises & Solutions

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction? $H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g)$ . Given that, bond energy of $H_{2}, {Br}_{2} and HBr$ is $435 kJ {mol}^{- 1}$ , $192 kJ {mol}^{- 1}$ and $368 kJ {mol}^{- 1}$ respectively.

Given that, bond energy of

H_{2} = 435 kJ {mol}^{- 1}

bond energy of

{Br}_{2} = 192 kJ {mol}^{- 1}

bond energy of

HBr = 368 kJ {mol}^{- 1}

For the reaction

H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g)

∆_{r} H^{⊝} = \sum BE (Reactants) - \sum BE (Products)

= BE (H_{2}) + BE ({Br}_{2}) - 2 BE (HBr)

= 435 + 192 - (2 \times 368) kJ {mol}^{- 1} .

= - 109 kJ {mol}^{- 1}

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