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Chemistry Question 71 – JEE-MAIN 2025

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Given : ΔHsub⊖[C (graphite)]=710 kJ mol−1 ΔC-HH⊖=414 kJ mol−1 ΔH-HH⊖=436 kJ mol−1 ΔC=CH⊖=611 kJ mol−1 The ΔHf⊖ for CH2=CH2 is _______ kJ mol−1 (nearest integer value)

The enthalpy of formation of a compound can be estimated using the bond enthalpies of the reactants and products involved in its formation reaction.

Step 1: Write the formation reaction and identify bonds✦ Active

The standard formation reaction for ethene (CH2=CH2) is:

2C (graphite)+2H2(g)→CH2=CH2(g)

To use bond enthalpies, we consider the atomization of reactants and then the formation of bonds in the product. The product, ethene, contains one C=C bond and four C-H bonds.

Step 2: Apply the bond enthalpy formula○ Expand

The enthalpy of formation (ΔHf⊖) can be calculated as the sum of enthalpies of bonds broken in reactants minus the sum of enthalpies of bonds formed in products. For the formation reaction, this translates to:

ΔHf⊖=[2×ΔHsub⊖(C (graphite))+2×ΔH-HH⊖]−[4×ΔC-HH⊖+1×ΔC=CH⊖]
💡 Teacher's Secret Hint

Remember that bond breaking is endothermic (positive enthalpy) and bond formation is exothermic (negative enthalpy). The formula accounts for this by summing positive bond enthalpies for reactants and subtracting the sum of positive bond enthalpies for products.

Step 3: Substitute values and calculate○ Expand

Substitute the given values:

ΔHsub⊖[C (graphite)]=710 kJ mol−1 ΔC-HH⊖=414 kJ mol−1 ΔH-HH⊖=436 kJ mol−1 ΔC=CH⊖=611 kJ mol−1 ΔHf⊖=[2×710+2×436]−[4×414+1×611] ΔHf⊖=[1420+872]−[1656+611] ΔHf⊖=2292−2267 ΔHf⊖=25 kJ mol−1

The nearest integer value is 25.

💡 Teacher's Secret Hint

Double-check your arithmetic to avoid calculation errors.

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