MCQ
The enthalpy changes for the following processes are listed below :
$Cl_2(g) \to 2Cl(g), 242.3\, kJ\, mol^{-1}$
$I_2(g) \to 2I(g), 151.0\, kJ\, mol^{-1}$
$ICl(g) \to I(g) + Cl(g), 211.3 kJ\, mol^{-1}$
$I_2(s) \to  I_2(g), 62.76\, kJ\, mol^{-1}$
Given that the standard states for iodine and chlorine are $I_2(s)$ and $Cl_2(g)$, the standard enthalpy of formation for $ICl(g)$ is .............. $\mathrm{kJ\,mol}^{-1}$
  • $+16.8$
  • B
    $+244.8$
  • C
    $-14.6$
  • D
    $-16.8$

Answer

Correct option: A.
$+16.8$
a
${I_2}(s) + C{l_2}(g) \to 2ICl(g)$

${\Delta _r}H = [\Delta H({I_2}(s) \to {I_2}(g)) + \Delta {H_{I - I}}$ $ + \Delta {H_{Cl - Cl}}] - [\Delta {H_{I - Cl}}]$

$ = 151.0 + 242.3 + 62.76 - 2 \times 211.3 = 33.46$

${\Delta _f}{H^o}(ICl) = \frac{{33.46}}{2} = 16.73\,kJ/mol$

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