Exams › JEE Main › Chemistry
Given: ΔsH°b [C(graphite)] = 710 kJ mol−1 ΔcH°e = 414 kJ mol−1 ΔhH°e = 436 kJ mol−1 ΔcH°e = 611 kJ mol−1 The ΔfH° for CH2=CH2 is ____ kJ mol−1 (nearest integer value)
- 125
- 126
- 124
- 127
Correct answer: 125
Solution
The correct option is derived from Hess's law, which allows us to calculate the standard enthalpy of formation (ΔfH°) for ethylene (CH2=CH2) by using the given enthalpy values for combustion and bond formation. By applying the appropriate equations and combining the enthalpy changes, we arrive at the value of 125 kJ mol−1.
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