The enthalpy change for the following reaction is 368 kJ. Calculate the average O–F bond…
Chemistry · JEE Main · NTA Exams — Chemical Thermodynamics
The enthalpy change for the following reaction is 368 kJ. Calculate the average O–F bond energy.
\(\mathrm{OF}_{2}(\mathrm{~g}) \rightarrow \mathrm{O}(\mathrm{~g})+2 \mathrm{~F}(\mathrm{~g})\)
\(\mathrm{OF}_{2}(\mathrm{~g}) \rightarrow \mathrm{O}(\mathrm{~g})+2 \mathrm{~F}(\mathrm{~g})\)
- 184 kJ/mol
- 368 kJ/mol
- 536 kJ/mol
- 736 kJ/mol
Answer
(A) 184 kJ/mol
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