Question

If the bond energies of \[H - H,Br - Br{\text{ and }}H - Br\] are \[433,192{\text{ and }}364{\text{ KJ mo}}{{\text{l}}^{ - 1}}\] respectively, the \[\Delta {H^0}\] for the reaction \[{H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}\] is:
A. \[ - 261{\text{ KJ}}\]
B. \[{\text{ + 103 KJ}}\]
C. \[{\text{ + 261 KJ}}\]
D. \[ - 103{\text{ KJ}}\]

Answer 1

Hint: First of all, we will calculate the values of bond enthalpies of \[H - H,Br - Br{\text{ and }}H - Br\] with positive or negative sign according to their bond energies and add up them to form the given reaction and hence to calculate the \[\Delta {H^0}\] for the given reaction.

Complete step-by-step answer:
Given reaction is \[{H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}\]
Given bond enthalpy of \[H - H = 433{\text{ KJ mo}}{{\text{l}}^{ - 1}}\]
Bond enthalpy of \[Br - Br = 192{\text{ KJ mo}}{{\text{l}}^{ - 1}}\]
Bond enthalpy of \[H - Br = 364{\text{ KJ mo}}{{\text{l}}^{ - 1}}\]
Reaction for the given bond enthalpies are given by
\[{H_{2\left( g \right)}} \to 2{H_{\left( g \right)}};\Delta H = 433{\text{ KJ mo}}{{\text{l}}^{ - 1}}\] (its value is positive because it is the energy required to break the bond)
\[B{r_{2\left( g \right)}} \to 2B{r_{\left( g \right)}};\Delta H = 192{\text{ KJ mo}}{{\text{l}}^{ - 1}}\] (its value is positive because it is the energy required to break the bond)
\[{H_{\left( g \right)}} + B{r_{\left( g \right)}} \to HB{r_{\left( g \right)}};\Delta H = - 364{\text{ KJ mo}}{{\text{l}}^{ - 1}}\] (its value is negative because it is the energy released due to bond formation)
We multiply third equation with 2 and then add three equation to get to the equation
\[{H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}\] which is the given reaction.
Hence the enthalpy change for the reaction is given by
\[
  \Delta {H^0} = 433 + 192 + 2\left( { - 364} \right) \\
  \therefore \Delta {H^0} = - 103{\text{ KJ }} \\
\]
Therefore \[\Delta {H^0}\] for the reaction \[{H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}\] is \[ - 103{\text{ KJ}}\]
Thus, the correct option is D. \[ - 103{\text{ KJ}}\]

Note: The change in enthalpy is defined as the amount of energy that is emitted off or absorbed into the system when a substantial mole of the reactants reacts to form one mole of products. The bond energy of a compound is the amount of energy that is released when the bond is formed between two atoms.