Question

The sublimation energy of ${I_2}\,(s)$ is $57.3\,kJ\,mo{l^{ - 1}}$ and the enthalpy of fusion is $15.5\,kJ\,mo{l^{ - 1}}$ . The enthalpy of evaporation of ${I_2}\,(s)$ is:
A) $41.8\,kJ\,mo{l^{ - 1}}$
B) $ - 41.8\,\,kJ\,mo{l^{ - 1}}$
C) $72.8\,\,kJ\,mo{l^{ - 1}}$
D) $ - 72.8\,\,kJ\,mo{l^{ - 1}}$

Answer 1

Hint:Write both values for the equation and calculate as by adding or subtracting both of the equation to get the final one. Final one is the evaporation of iodine that the final equation will be the conversion of solid iodine into vapours. For completing the whole equation you have to balance it and write the enthalpy of the reaction.

Complete solution:
We have given that the sublimation energy of iodine is $57.3\,J\,mo{l^{ - 1}}$ it means the conversion of solid iodine directly into the vapours or gaseous form requires this amount of energy. Similarly for the other part of the question that is the enthalpy for the fusion of iodine means conversion of solid iodine into the liquid form is given as $15.5\,J\,mo{l^{ - 1}}$ value.
Now if we write it in the equation form it will be just like this.
${I_2}\,(s) \to \,{I_2}\,(g)\,\,\,\,\,\,\,\Delta H = \,57.3kJ\,mo{l^{ - 1}}$
${I_2}\,(s) \to \,{I_2}\,(l)\,\,\,\,\,\,\,\Delta H = \,15.5kJ\,mo{l^{ - 1}}$
Let the first equation be number $(1)$ and the second equation as $(2)$ now we want to calculate the enthalpy for evaporation that means when iodine which is actually present in liquid form changes to vapours. ${I_2}\,(l) \to \,{I_2}\,(g)\,\,$
Now for achieving this equation, we have to reverse equation $(2)$ and then we will get this type of equation: ${I_2}\,(l) \to \,{I_2}\,(s)\,\,\,\,\,\,\,\Delta H = \, - 15.5kJ\,mo{l^{ - 1}}$ write it as equation $(3)$
Now let’s add both equation by which we will get equation like this: ${I_2}\,(l)\, + \,{I_2}\,(s) \to \,{I_2}\,(g)\, + \,{I_2}\,(s)\,$
Solid phase iodine gets canceled from both sides and we get our required equation ${I_2}\,(l) \to \,{I_2}\,(g)\,\,$ that is equation$(3)$ . For finding the value of enthalpy we have to add both enthalpies of equations and then we will get the enthalpy for the equation $(3)$ .
$\Delta {H_3} = \,\Delta {H_1}\, + \,\Delta {H_2}$
$\Delta {H_3} = \,(57.3 - \,15.5)\,kJ\,mo{l^{ - 1}}$
$\Delta {H_3} = \,41.8\,\,kJ\,mo{l^{ - 1}}$
Therefore, option A. is correct.

Note:We have to change the sign in enthalpy when you are reversing the equation $(2)$ . It conveys that the reaction of converting solid iodine into its liquid form gets reversed from liquid iodine to solid iodine. Thus if $15.5\,J\,mo{l^{ - 1}}$ amount of energy is required for converting solid iodine into liquid, then for its opposite conversion the value becomes negative.