The energy absorbed by each molecule (A_2) of | vedclass

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(D) Energy absorbed by molecule $(A_2) = 4.4 	imes 10^{-19} \ J$ Bond energy per molecule $(A_2) = 4.0 	imes 10^{-19} \ J$ Excess energy available a

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$2.0$

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(D) Energy absorbed by molecule $(A_2) = 4.4 	imes 10^{-19} \ J$ Bond energy per molecule $(A_2) = 4.0 	imes 10^{-19} \ J$ Excess energy available as kinetic energy $(K.E.) = 	ext{Energy absorbed} - 	ext{Bond energy}$ $K.E. = 4.4 	imes 10^{-19} - 4.0 	imes 10^{-19} = 0.4 	imes 10^{-19} \ J$ Since the molecule is $(A_2)$,it contains $2$ atoms. $K.E.$ per atom $= \frac{0.4 	imes 10^{-19} \ J}{2} = 0.2 	imes 10^{-19} \ J = 2 	imes 10^{-20} \ J$ Thus,the kinetic energy per atom is $2 	imes 10^{-20} \ J$.

Molecule	$\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$	$-280$
$CO_{2(g)}$	$-400$
$H_2O_{(l)}$	$-290$

The energy absorbed by each molecule $(A_2)$ of a substance is $4.4 \times 10^{-19} \ J$ and bond energy per molecule is $4.0 \times 10^{-19} \ J.$ The kinetic energy of the molecule per atom will be $...... \times 10^{-20} \ J$

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