Question

Consider the reaction at $300\, K$
$H _{2( g )}+ Cl _{2( g )} \rightarrow 2 HCl _{( g )}, \Delta H ^{\circ}=-185\, kJ \,mol ^{-1}$
If $3$ mole of $H _{2}$ completely react with $3 \,mol$ of $Cl _{2}$ to form $HCl , \Delta U ^{\circ}$ of the reaction will be :

• A. Zero
• B. - 185 kJ
• C. -555 kJ
• D. - 200 kJ

Answer 1

Answer: (C)

$H _{2}+ Cl _{2} \longrightarrow 2 HCl \,\,\,\,\,\Delta H =-185\, kJ$
for $3$ mole $H _{2} \,\&\, 3$ mole $Cl _{2}$
$\Delta H =-185 \times 3=-555 \,kJ $
$\Delta U =\Delta H$