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)

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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