JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
For the reaction \(C _{2} H _{6} \rightarrow C _{2} H _{4}+ H _{2}\) the reaction enthalpy \(\Delta_{ r } H =...........{ kJ\, mol ^{-1}}\). (Round off to the Nearest Integer). [Given : Bond enthalpies in \(kJ\) \(mol\) \(^{-1}:C-C : 347, C = C : 611 ; C - H : 414, H - H : 436]\)
- A \(120\)
- B \(132\)
- C \(128\)
- D \(125\)
Answer & Solution
Correct Answer
(C) \(128\)
Step-by-step Solution
Detailed explanation
\(\Delta_{ r } H =\left[\epsilon_{ C - C }+2 \epsilon_{ C - H }\right]-\left[\epsilon_{ C = C }+\epsilon_{ H - H }\right]\) \(=[347+2 \times 414]-[611+436]\) \(=128\)
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