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Question: The standard heat of the combustion of ethene gas is -330K \[calmo{l^{ - 1}}\]. Calculate C=C bond e...

The standard heat of the combustion of ethene gas is -330K calmol−1calmo{l^{ - 1}}. Calculate C=C bond energy (in kcal/mol) assuming that the bond energy of C-H bond is 93.6 K calmol-1.
Given that ΔHf∘forCO2\Delta H_f^ \circ forC{O_2} and H2O{H_2}O are -94.2 and -61K calmol-1 respectively. Heat of atomization of carbon and hydrogen are 150 and 51.5 K calmol-1 respectively.
A. 82 K cal/mol
B. 92 K cal/mol
C. 102 K cal/mol
D. 112 K cal/mol

Explanation

Solution

We know that whenever a chemical bond is formed energy is released and conversely energy required to break a bond is bond dissociation energy. bond energy can be expressed as the average of bond dissociation energies of various similar bonds.

Complete step by step answer:
We are given with the standard heat of the combustion of ethene gas as -330K and heat of atomization which is the amount of heat required for the formation of one mole of atoms in gaseous state from its elements for carbon and hydrogen. Let us write the reactions occurring,
CH2+3O2→2CO2+2H2O;ΔHC=−330Kcal ∥ CH2  C{H_2} + 3{O_2} \to 2C{O_2} + 2{H_2}O;\Delta {{\rm H}_C} = - 330Kcal \\\ \parallel \\\ C{H_2} \\\
2C+2O2→2CO2;ΔHf=−94.2Kcal/mol 2H2+O2→2H2O;ΔHf=−61Kcal/mol 2CO2+2H2O→C2H4+3O2;ΔHf=19.6Kcal/mol ΔHreaction=2×ΔHC+4×ΔHH−4B.EC−H−B.EC−C  2C + 2{O_2} \to 2C{O_2};\Delta {{\rm H}_f} = - 94.2Kcal/mol \\\ 2{H_2} + {O_2} \to 2{H_2}O;\Delta {{\rm H}_f} = - 61Kcal/mol \\\ 2C{O_2} + 2{H_2}O \to {C_2}{H_4} + 3{O_2};\Delta {{\rm H}_f} = 19.6Kcal/mol \\\ \Delta {{\rm H}_{reaction}} = 2 \times \Delta {{\rm H}_C} + 4 \times \Delta {{\rm H}_H} - 4B.{E_{C - H}} - B.{E_{C - C}} \\\
We know that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants and the standard enthalpy of formation values
Substituting the values, we get;
19.6 = 2 x 150 + 4 x 51.5-4 x 93.6 - BE(C=C)
BE(C=C) = 300+206-374.4-19.6
= 506-394=112 Kcal/mol

Hence we can conclude that option D is correct.

Note:
The amount of heat released for complete combustion of a compound in its standard state to form stable compounds in their standard states. The factors which affect the heat of combustion is temperature, since it releases the energy it is always exothermic, the enthalpy change for the reaction is negative.