Question

Using the following data, calculate the standard heat of formation of ICl(g) in kJ/mol:

$C{l}_2\left(g\right)\to 2C\text{}l\left(g\right)\Delta H=242.3kJ$

$l_2\left(g\right)\to 2\text{}l\left(g\right)\Delta H=151kJ$

$lCl\left(g\right)\to \text{\hspace{0.17em}}l\left(g\right)+Cl\left(g\right)\Delta H=211.3kJ$

$l_2\left(s\right)\to \text{\hspace{0.17em}}{l}_2\left(g\right)\Delta H=62.8kJ$

Short answer

Standard Heat of Formation of ICl(g) is 16.75 kJ.

Step-by-step solution

Standard Heat of Formation

Standard heat of formation givesthe amount of heat which is absorbed or evolved when one mole of a compound is formed from its elements, each of substance should be in its normal physical state i.e., gas, liquid, or solid.

Thus, standard heat for formation of ICl can be shown as

$\frac{1}{2}\text{\hspace{0.17em}}{l}_2\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to lCl\left(g\right)\Delta H=standardheatofformation$

 Step 2: Steps Involved in Formation of ICl

First step is the conversion of I₂ solid to I₂ gas i.e., sublimation of I₂:

$l_2\left(s\right)\to \text{\hspace{0.17em}}{l}_2\left(g\right)\Delta H=62.8kJ$

$\frac{1}{2}{l}_2\left(s\right)\to \frac{1}{2}\text{\hspace{0.17em}}{l}_2\left(g\right)\Delta H=\frac{62.8}{2}=31.4kJ\mathrm{.....}\left(i\right)$

Second step is the conversion of I₂ gas to atoms of I(g) i.e., atomization

$l_2\left(g\right)\to 2\text{}l\left(g\right)\Delta H=151kJ$

Energy required to form one mole I(g) is

$\frac{1}{2}{l}_2\left(g\right)\to \text{\hspace{0.17em}}l\left(g\right)\Delta H=\frac{151}{2}=75.5kJ\mathrm{.....}\left(ii\right)$

Third step is the conversion of Cl₂ gas to atoms of Cl(g) i.e., atomization

$C{l}_2\left(g\right)\to 2C\text{}l\left(g\right)\Delta H=242.3kJ$

Energy required to form one mole Cl(g) is

$\frac{\mathbf{1}}{\mathbf{2}}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{C}\mathbf{\text{}}\mathit{l}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{\Delta }\mathit{H}\mathbf{=}\frac{\mathbf{242}\mathbf{.}\mathbf{3}}{\mathbf{2}}\mathbf{=}\mathbf{121}\mathbf{.}\mathbf{15}\mathit{k}\mathit{J}\mathbf{.....}\mathbf{\left(}\mathit{i}\mathit{i}\mathit{i}\mathbf{\right)}$

Fourth step is the reaction of I(g) and Cl(g) gas to form ICl(g) i.e., bond formation.

$\frac{1}{2}\text{\hspace{0.17em}}{l}_2\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to lCl\left(g\right)\Delta H=16.75kJ$

Calculating Standard Heat of Formation of ICl Using the Steps Involved

By adding all (i) to (iv) we can get standard heat of formation so,

$\frac{1}{2}\text{\hspace{0.17em}}{l}_2\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to lCl\left(g\right)\Delta H=16.75kJ$