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Question: Enthalpy of sublimation of iodine is 24 \( cal{{g}^{-1}}\) at 200\(^{\circ }C\). If the specific hea...

Enthalpy of sublimation of iodine is 24 calg−1 cal{{g}^{-1}} at 200∘C^{\circ }C. If the specific heat of I2{{I}_{2}}(s) and I2{{I}_{2}}(vap) are 0.055 and 0.031calg−1K−1cal{{g}^{-1}}{{K}^{-1}} respectively, then enthalpy of sublimation of iodine at 250∘C^{\circ }C in calg−1 cal{{g}^{-1}}is :
(a) 2.85
(b) 11.4
(c) 5.7
(d) 22.8

Explanation

Solution

. First we have to calculate the change in the specific heat capacity of reactants and products and the change in the temperature by using the given data and then we can easily calculate the enthalpy of sublimation by using the formula as ΔH=ΔcP(T2−T1)\Delta H=\Delta {{c}_{P}}({{T}_{2}}-{{T}_{1}}). Now solve it.

Complete step by step answer:
By the enthalpy of sublimation, we mean the amount of energy that is required to change the substance from its solid phase to its vapor phase.
Now, considering the numerical:
Consider the reaction:
I2(s)→I2(v){{I}_{2}}(s)\to {{I}_{2}}(v)

We can calculate the enthalpy of sublimation of iodine by using the formula as;
ΔH=ΔcP(T2−T1)\Delta H=\Delta {{c}_{P}}({{T}_{2}}-{{T}_{1}}) ------------(1)
We can calculate the change in the specific heat as we have been the specific heat capacities of I2{{I}_{2}}(s) and I2{{I}_{2}}(vap).
Specific heat capacity of I2{{I}_{2}}(s)= 0.055 calg−1K−1cal{{g}^{-1}}{{K}^{-1}}
Specific heat capacity of I2{{I}_{2}}(vap)= 0.031 calg−1K−1cal{{g}^{-1}}{{K}^{-1}}

Then , the change in the specific heat capacity is as;
ΔcP=cp(product)−cp(reactants)\Delta {{c}_{P}}={{c}_{p(product)}}-{{c}_{p(reactants)}}
= 0.031-0.055
= -0.024calg−1cal{{g}^{-1}}
Temperature ,T1{{T}_{1}}= 200∘C^{\circ }C
= 200 +273 K
= 473K
Temperature, T2{{T}_{2}}=250∘C^{\circ }C
= 250 +273 K
= 523K
Then ,the change in temperature is as:
ΔT=T2−T1\Delta T={{T}_{2}}-{{T}_{1}}
= 523- 473K
= 50K
And we know that ΔH1\Delta {{H}_{1}}=24
ΔH=ΔH2−ΔH1\Delta H = \Delta {{H}_{2}}-\Delta {{H}_{1}}
=ΔH2−24\Delta {{H}_{2}}-24

Then , put all these values in equation(1), we get;
ΔH2−24=−0.024(50)\Delta {{H}_{2}}-24 = -0.024(50)
ΔH2\Delta {{H}_{2}} = -1.2 + 24
= 22.8 calg−1cal{{g}^{-1}}
Hence the enthalpy of sublimation of iodine at 250∘C^{\circ }C is : 22.8 calg−1cal{{g}^{-1}}
So, the correct answer is “Option D”.

Note: Temperature should always be taken in the standard units i.e. the Kelvin and if the temperature is in degree Celsius convert it into Kelvin by adding 273 to the given temperature in degree Celsius.