Question

The enthalpy changes for the following processes are listed below:
$C{l_{2(g)}} \to 2C{l_{(g)}};242.3KJmo{l^{ - 1}}$
${I_{2(g)}} \to 2{I_{(g)}};151.0KJmo{l^{ - 1}}$
$IC{l_{(g)}} \to {I_{(g)}} + C{l_{(g)}};211.3KJmo{l^{ - 1}}$
${I_2}_{(s)} \to {I_{2(g)}};62.57KJmo{l^{ - 1}}$
Given that the standard states for iodine and chlorine are ${I_{2(s)}}$ and $C{l_{2(g)}}$, the standard enthalpy of formation $IC{l_{(g)}}$ is:
(A) $ - 16.8KJmo{l^{ - 1}}$
(B) $ + 16.8KJmo{l^{ - 1}}$
(C) $ + 244.8KJmo{l^{ - 1}}$
(D) $ - 14.6KJmo{l^{ - 1}}$

Answer 1

Hint:We have the formula for the standard enthalpy change as
Standard enthalpy change $ = $ (Sum of standard enthalpy of formation of products $ - $ Sum of standard enthalpy of formation of reactants)
The standard enthalpy of a reaction is thus the change of reactants to products.

Complete step by step answer:
Let us first understand what is the standard enthalpy of formation.
It has been even named as standard heat of formation.
As the word defines itself, it is the change in enthalpy or energy when one mole of a substance is formed from its constituent elements. It is usually measured in kilojoule per mole or joule per mole.
Now, coming to the question, we have been given
The standard enthalpy of formation of $2C{l_{(g)}} = 242.3KJmo{l^{ - 1}}$
The standard enthalpy of formation of $2{I_{(g)}} = 151.0KJmo{l^{ - 1}}$
The standard enthalpy of formation of ${I_{(g)}} + C{l_{(g)}} = 211.3KJmo{l^{ - 1}}$
The standard enthalpy of formation of ${I_{2(g)}} = 62.75KJmo{l^{ - 1}}$
We only one mole of $IC{l_{(g)}}$ so we take half a mole of ${I_{2(g)}}$ and half a mole of $C{l_{2(g)}}$
$\dfrac{1}{2}{I_{2(s)}} + \dfrac{1}{2}C{l_{2(g)}} \to IC{l_{(g)}}$
The following equation can be used to calculate the standard enthalpy of reaction:
             $\sum {\Delta H_{rxn}^ - } = \sum {\Delta H_f^ - } [products] - \sum {\Delta H_f^ - [rct]} $
We change this reaction into half symmetrical
$C{l_{2(g)}} \to 2C{l_{(g)}}$
${I_{2(g)}} \to 2{I_{(g)}}$
$IC{l_{(g)}} \to {I_{(g)}} + C{l_{(g)}}$
${I_2}_{(s)} \to {I_{2(g)}}$
$\Delta {H_f} = \dfrac{1}{2}\left( {243.5 + 151.0 + 62.76} \right) - 211.3$
$ = + 16.8kJ/mol$
Thus, the value for the standard enthalpy change comes out to be $ + 16.8KJmo{l^{ - 1}}$.

Thus, the correct answer is (B).

Note:
It must be noted that change can be positive or negative. If the energy is consumed in the reaction then the value comes out to be positive and the reaction in nature and causes the heat is released then the value comes out to be negative and is an exothermic reaction.
The units while subtraction should be the same.
Further, the reactant and product values of each type of molecule are to be multiplied by coefficients present in a balanced reaction.