Chemical Thermodynamics — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryChemical Thermodynamics3 Marks
Question
How will you calculate reaction enthalpy from data on bond enthalpies?
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Answer
$(i)$ In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
$(ii)$ Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
$(iii)$ The enthalpy change of a gaseous reactions $( \Delta_{ f } H ^{ O } )$ involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. $($In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.$) \Delta H^0 ($ reaction $)=$
${\left[\begin{array}{c}\text { Sum of bond enthalpies } \\\text { of bonds broken } \\\text { of reactants }
\end{array}\right]-\left[\begin{array}{c}\text { Sum of bond } \\ \text { enthalpies of bonds } \\\text { formed of products }\end{array}\right]}$
$=\Sigma \Delta H_{\text {bonds broken }}^0-\Sigma \Delta H_{\text {bonds formed }}^0$
For example for a following reaction,
$\ce{H_{2(g)} + Cl_{2(g)} \longrightarrow 2HCl_{(g)}}$
$\ce{ORH-H_{(g)} + Cl-Cl_{(g)} \longrightarrow 2 H-Cl_{(g)}}$
$\ce{\Delta H_{reaction}^0=\left[\Delta H_{H-H}^0 + \Delta H_{C l-C l}^0\right]-2\left[\Delta H_{H-C l}^0\right]}$
If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and $\Delta H^0$ reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and $\Delta H^0$ reaction will be negative.
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