Looking for classes? Ksquare Career Institute, Bengaluru →
Bond dissociation enthalpy of $H_2$, $Cl_2$ and HCl are 434, 242 and 431 kJ mol$^{-1}$ respectively. Enthalpy of formation of HCl is -
A
$-93$ kJ mol$^{-1}$
B
245 kJ mol$^{-1}$
C
93 kJ mol$^{-1}$
D
$-245$ kJ mol$^{-1}$
Detailed Solution
$H_2 + Cl_2 \rightarrow 2HCl$
$\Delta H_{reaction} = \Sigma(BE)_{reactants} - \Sigma(BE)_{products}$
$= [(BE)_{H-H} + (BE)_{Cl-Cl}] - [2(BE)_{H-Cl}]$
$= 434 + 242 - (431) \times 2$
$= 676 - 862$
$= -186$ kJ
This is the enthalpy change for the formation of 2 moles of HCl.
So the enthalpy of formation of HCl (per mole) $= \dfrac{-186}{2}$ kJ
$= -93$ kJ mol$^{-1}$
$\Delta H_{reaction} = \Sigma(BE)_{reactants} - \Sigma(BE)_{products}$
$= [(BE)_{H-H} + (BE)_{Cl-Cl}] - [2(BE)_{H-Cl}]$
$= 434 + 242 - (431) \times 2$
$= 676 - 862$
$= -186$ kJ
This is the enthalpy change for the formation of 2 moles of HCl.
So the enthalpy of formation of HCl (per mole) $= \dfrac{-186}{2}$ kJ
$= -93$ kJ mol$^{-1}$