NEET · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
At standard conditions, if the change in the enthalpy for the following reaction is \(-109 \;kJ mol ^{-1}\) \(H _{2( g )}+ Br _{2( g )} \rightarrow 2 HBr _{( g )}\) Given that bond energy of \(H _{2}\) and \(Br _{2}\) is \(435 \;kJ mol ^{-1}\) and \(192 \;kJ mol^{-1}\), respectively, what is the bond energy (in \(kJ mol ^{-1}\) ) of \(HBr?\)
- A \(259\)
- B \(368\)
- C \(736\)
- D \(518\)
Answer & Solution
Correct Answer
(B) \(368\)
Step-by-step Solution
Detailed explanation
\(H _{2}( g )+ Br _{2}( g ) \longrightarrow 2 HBr ( g ) ; \Delta H =-109\) \(\Delta H =\sum( BE )_{ R }-\sum( BE )_{ P }\)
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