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Question: If the bond energies of \[H - H,Br - Br{\text{ and }}H - Br\] are \[433,192{\text{ and }}364{\text{ ...

If the bond energies of H−H,Br−Br and H−BrH - H,Br - Br{\text{ and }}H - Br are 433,192 and 364 KJ mol−1433,192{\text{ and }}364{\text{ KJ mo}}{{\text{l}}^{ - 1}} respectively, the ΔH0\Delta {H^0} for the reaction H2(g)+Br2(g)→2HBr(g){H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}} is:
A. −261 KJ - 261{\text{ KJ}}
B.  + 103 KJ{\text{ + 103 KJ}}
C.  + 261 KJ{\text{ + 261 KJ}}
D. −103 KJ - 103{\text{ KJ}}

Explanation

Solution

Hint: First of all, we will calculate the values of bond enthalpies of H−H,Br−Br and H−BrH - 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 ΔH0\Delta {H^0} for the given reaction.

Complete step-by-step answer:
Given reaction is H2(g)+Br2(g)→2HBr(g){H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}
Given bond enthalpy of H−H=433 KJ mol−1H - H = 433{\text{ KJ mo}}{{\text{l}}^{ - 1}}
Bond enthalpy of Br−Br=192 KJ mol−1Br - Br = 192{\text{ KJ mo}}{{\text{l}}^{ - 1}}
Bond enthalpy of H−Br=364 KJ mol−1H - Br = 364{\text{ KJ mo}}{{\text{l}}^{ - 1}}
Reaction for the given bond enthalpies are given by
H2(g)→2H(g);ΔH=433 KJ mol−1{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)
Br2(g)→2Br(g);ΔH=192 KJ mol−1B{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(g)+Br(g)→HBr(g);ΔH=−364 KJ mol−1{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
H2(g)+Br2(g)→2HBr(g){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

ΔH0=433+192+2(−364) ∴ΔH0=−103 KJ   \Delta {H^0} = 433 + 192 + 2\left( { - 364} \right) \\\ \therefore \Delta {H^0} = - 103{\text{ KJ }} \\\

Therefore ΔH0\Delta {H^0} for the reaction H2(g)+Br2(g)→2HBr(g){H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}} is −103 KJ - 103{\text{ KJ}}
Thus, the correct option is D. −103 KJ - 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.