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  • At standard conditions, if the change in the enthalpy for the following reaction is

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 2HBr__{(g)}

Given that bond energy of H_2 and Br_2 is 435 kJ mol^{-1}\ and\ 192 kJmol^{-1} respectively, What is the bond energy (in kJ mol-1) of HBr?

Option: 1

736


Option: 2

518


Option: 3

259


Option: 4

368


Answers (1)

best_answer

H_2\rightarrow 2H\quad B.E_{H_2}\\ Br_2\rightarrow 2Br \quad B.E_{Br_2}\\ 2H+2Br\rightarrow 2HBr\quad -2B.E_{HBr}\\ H_2+Br_2\rightarrow 2HBr

From the Bond energy concept, we have

BE_{H_2}+BE_{Br_2} - 2BE_{HBr}=\Delta H_{HBr}\\ \rightarrow 435+192-2x=-109\\ \rightarrow 2x=736\\\rightarrow x=368\: kJ\: mol^{-1}

Hence the correct answer is option 4

Posted by

Rishabh

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