Solveeit Logo
Login

Question

Question: The sublimation energy of \({I_2}\,(s)\) is \(57.3\,kJ\,mo{l^{ - 1}}\) and the enthalpy of fusion is...

The sublimation energy of I2 (s){I_2}\,(s) is 57.3 kJ mol−157.3\,kJ\,mo{l^{ - 1}} and the enthalpy of fusion is 15.5 kJ mol−115.5\,kJ\,mo{l^{ - 1}} . The enthalpy of evaporation of I2 (s){I_2}\,(s) is:
A) 41.8 kJ mol−141.8\,kJ\,mo{l^{ - 1}}
B) −41.8  kJ mol−1 - 41.8\,\,kJ\,mo{l^{ - 1}}
C) 72.8  kJ mol−172.8\,\,kJ\,mo{l^{ - 1}}
D) −72.8  kJ mol−1 - 72.8\,\,kJ\,mo{l^{ - 1}}

Explanation

Solution

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 mol−157.3\,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 mol−115.5\,J\,mo{l^{ - 1}} value.
Now if we write it in the equation form it will be just like this.
I2 (s)→ I2 (g)       ΔH= 57.3kJ mol−1{I_2}\,(s) \to \,{I_2}\,(g)\,\,\,\,\,\,\,\Delta H = \,57.3kJ\,mo{l^{ - 1}}
I2 (s)→ I2 (l)       ΔH= 15.5kJ mol−1{I_2}\,(s) \to \,{I_2}\,(l)\,\,\,\,\,\,\,\Delta H = \,15.5kJ\,mo{l^{ - 1}}
Let the first equation be number (1)(1) and the second equation as (2)(2) now we want to calculate the enthalpy for evaporation that means when iodine which is actually present in liquid form changes to vapours. I2 (l)→ I2 (g)  {I_2}\,(l) \to \,{I_2}\,(g)\,\,
Now for achieving this equation, we have to reverse equation (2)(2) and then we will get this type of equation: I2 (l)→ I2 (s)       ΔH= −15.5kJ mol−1{I_2}\,(l) \to \,{I_2}\,(s)\,\,\,\,\,\,\,\Delta H = \, - 15.5kJ\,mo{l^{ - 1}} write it as equation (3)(3)
Now let’s add both equation by which we will get equation like this: I2 (l) + I2 (s)→ I2 (g) + I2 (s) {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 I2 (l)→ I2 (g)  {I_2}\,(l) \to \,{I_2}\,(g)\,\, that is equation(3)(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)(3) .
ΔH3= ΔH1 + ΔH2\Delta {H_3} = \,\Delta {H_1}\, + \,\Delta {H_2}
ΔH3= (57.3− 15.5) kJ mol−1\Delta {H_3} = \,(57.3 - \,15.5)\,kJ\,mo{l^{ - 1}}
ΔH3= 41.8  kJ mol−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)(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 mol−115.5\,J\,mo{l^{ - 1}} amount of energy is required for converting solid iodine into liquid, then for its opposite conversion the value becomes negative.