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  • The enthalpy changes for the following processes are listed below: <img alt="\small Cl_{2(g)}=2Cl_{g},242.3kJ\: mol^{-1}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%

The enthalpy changes for the following processes are listed below:

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

\small I_{2(g)}=2I_{g},151.0kJ\: mol^{-1}

\small ICl_{(g)}=I_{g}+Cl_{(g)},211.3kJ\: mol^{-1}

\small I_{2(s)}=I_{2(g)},62.76kJ\: mol^{-1}

Given that the standard states for iodine and chlorine are \small I_2 and \small Cl_{2(g)} , the standard enthalpy of formation for  \small ICl_{(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 kJ mol^{-1}


Answers (1)

best_answer

The state of the reactant needs to change in the gaseous state.

\mathrm{\frac{1}{2}\times I_{2(s)}+\frac{1}{2}\times Cl_{2(g)}\rightarrow ICl_{(g)}}

\\\mathrm{\Delta H=[2\Delta H_{I-Cl}]-[\Delta H_{I_2(s)\rightarrow I_{2(g)}}+\Delta H_{I-I}+\Delta H_{Cl-Cl}]}\\\mathrm{\Delta H=[\frac{1}{2}\times\Delta H_{I_2(s)\rightarrow I_{2(g)}}+\frac{1}{2}\times\Delta H_{I-I}+\frac{1}{2}\times\Delta H_{Cl-Cl}]-[\Delta H_{I-Cl}]}

\\\mathrm{\Delta H =(\frac{1}{2}\times62.76+\frac{1}{2}\times151.0+\frac{1}{2}\times242.3)- 211.3} \\\\\mathrm{\Delta H =16.73\ kJ/mol}

Therefore, option(3) is correct.

Posted by

Deependra Verma

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