Question

From the following bond energies: H - H bond energy: {E_0} = {{ }}2679.47 - 2799.49{{ }} = {{ }} - 120.02{{ }}kJ{{ }}mo{l^{ - 1}}$$$$431.37{{ }}kJ{{ }}mo{l^{ - 1}} , C = C bond energy: $606.10{{ }}kJ{{ }}mo{l^{ - 1}}$ , C - C bond energy: $336.49{{ }}kJ{{ }}mo{l^{ - 1}}$ , C - H bond energy: $410.50{{ }}kJ{{ }}mo{l^{ - 1}}$ . Enthalpy for the reaction will be:
A. $- {{ }}243.6{{ }}kJ{{ }}mo{l^{ - 1}}$
B. $- 120.0{{ }}kJ{{ }}mo{l^{ - 1}}$
C. $553.0{{ }}kJ{{ }}mo{l^{ - 1}}$
D. $1523.6{{ }}kJ{{ }}mo{l^{ - 1}}$

Answer 1

We know that the overall enthalpy for any reaction can be found out by using the formula given below:
${E_O} = {E_R} - {E_P}$
Where, ${E_O}$ is the enthalpy of the overall given reaction,
${E_R}$ is the bond dissociation energy of reactant,
${E_P}$ is the bond dissociation energy of the product.

Complete step by step answer:
The reaction given in the question statement is an example of the simplest reduction reaction in which when hydrogen gas reacts with an alkene (ethene in this case) it leads to the formation of alkane (ethane in this case).
In the question statement, we are provided by:
H - H bond energy: $B.E{._{H - H}} = {{ }}431.37{{ }}kJ{{ }}mo{l^{ - 1}}$
C = C bond energy: $B.E{._{C = C}} = {{ }}606.10{{ }}kJ{{ }}mo{l^{ - 1}}$
C - C bond energy: $B.E{._{C - C}} = {{ }}336.49{{ }}kJ{{ }}mo{l^{ - 1}}$
C - H bond energy: $B.E{._{C - H}} = {{ }}410.50{{ }}kJ{{ }}mo{l^{ - 1}}$
So, now if we look at the reaction carefully, there are four C-H single bonds, one C=C single bond in ethene molecule, and one H-H single in hydrogen gas molecule. In the product, we have six C-H single bond and C-C single bond in the ethane molecule. Using this data, we will find the enthalpy of the overall reaction by using the formula given below:
${E_O} = {E_R} - {E_P}$
Where, ${E_O}$ is the enthalpy of the overall given reaction,
${E_R}$ is the bond dissociation energy of reactant,
${E_P}$ is the bond dissociation energy of the product.
Now, from the reaction, we know that the bond dissociation energy of the reactant will correspond to the sum of bond dissociation energies of four C-H, one C=C, and one H-H bond.
Thus, ${E_R} = {{ }}4B.E{._{C - H}} + {{ }}B.E{._{C = C}} + {{ }}B.E{._{H - H}}$
And, bond dissociation energy of the product will correspond to the sum of bond dissociation energies of six C-H and one C-C bond.
Thus, ${E_P} = {{ }}6{{ }}B.E{._{C - H}} + {{ }}B.E{._{C - C}}$
Therefore, the enthalpy of the overall reaction would be:
${E_0} = {{ }}\left( {4B.E{._{C - H}} + {{ }}B.E{._{C = C}} + {{ }}B.E{._{H - H}}} \right){{ }} - \left( {{{ }}6{{ }}B.E{._{C - H}} + {{ }}B.E{._{C - C}}} \right)$
${E_0} = {{ }}\left[ {606.1 + \left( {4 \times 410.5} \right){{ }} + 431.37} \right]{{ }}-{{ }}\left[ {6 \times 410.5 + 336.49} \right]$
${E_0} = {{ }}\left( {606.1 + 1642 + 431.37} \right){{ }}-{{ }}\left( {2463 + 336.49} \right)$
${E_0} = {{ }}2679.47 - 2799.49{{ }} = {{ }} - 120.02{{ }}kJ{{ }}mo{l^{ - 1}}$
Hence. The correct option is option B. the enthalpy of the given reaction is $- 120.02{{ }}kJ{{ }}mo{l^{ - 1}}$

So, the correct answer is Option B.

Note: Here in this reaction, we noticed that the enthalpy value of the product is lower than the enthalpy value of the reactants. Due to this, the overall enthalpy of reaction came out to be negative. This suggests that the given reaction is exothermic, a reaction during which heat is being released in the surrounding.