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The enthalpy changes for the following processes are listed below:

$Cl_{2(g)}=2Cl_{(g)},242.3\, kJ\; mol^{-1}$

$l_{2(g)}=2l_{(g)},151.0\, kJ\; mol^{-1}$

$lCl_{(g)}=l_{(g)}+Cl_{(g)},211.3\, kJ\; mol^{-1}$

$l_{2(s)}=l_{2(g)},62.76\, kJ\; mol^{-1}$

Given that the standard states for iodine and chlorine are $l_{2(s)}$ and $Cl_{2(g)}$ , the standard enthalpy of formation for $lCl_{(g)}$ is

Option 1)
– 14.6 kJ mol^-1

Option 2)
– 16.8 kJ mol^-1

Option 3)
+ 16.8 kJ mol^-1

Option 4)
+ 244.8 kJ mol^-1

Answers (1)

best_answer

As we learnt in

Hess's Law -

The total amount of heat change in a chemical reaction is the same whether the reaction is carried out in one or several steps by one or more methods.

- wherein

$C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}$

$\Delta H= 393.5\, kj$

$C_{(s)}+\frac{1}{2}O_{2(g)}\rightarrow CO_{(g)}$

$\Delta H_{1}= 110.5\, kJ$

$CO_{(g)}+\frac{1}{2}O_{2(g)}\rightarrow CO_{2(g)}$

$\Delta H_{2}= 283.0\, kJ$

$\Delta H= \Delta H_{1}+\Delta H_{2}$

$I_{2}(S)+Cl_{2}(g)\rightarrow 2I\:Cl(g)$

$\Delta H\:=\:[\Delta H_{I_{2}(S)\rightarrow I_{2}g}+\Delta H_{I-I}+\Delta H_{Cl-Cl}]-[\Delta H_{I-Cl}]$

$151.0+242.3+62.76-2\times211.3\:=33.46$

$\Delta H^{\circ}_{F}(ICl)\:=\frac{33.46}{2}=16.73KJ/mol$

Option 1)

– 14.6 kJ mol^-1

This option is incorrect.

Option 2)

– 16.8 kJ mol^-1

This option is incorrect.

Option 3)

+ 16.8 kJ mol^-1

This option is correct.

Option 4)

+ 244.8 kJ mol^-1

This option is incorrect.

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