ThermochemistryHard
Question
The specific heats of iodine vapours and solid are 0.031 and 0.055 cal/g, respectively. If the enthalpy of sublimation of iodine is 24 cal/g at 200oC, then the enthalpy of sublimation of iodine at 250o C should be
Options
A.24 cal/g
B.22.8 cal/g
C.26.4 cal/g
D.25.2 cal/g
Solution
$I_{2}(s) \rightarrow I_{2}(g);$
$\Delta H_{1} = 24\text{ cal/gmat }\text{T}_{1} = 473\text{ K} $$$\Delta H_{2} = ? \text{at }\text{T}_{2} = 523\text{ K}$$
$\frac{\Delta H_{2} - \Delta H_{1}}{T_{2} - T_{1}} = C_{p},I_{2}(g) - C_{p},I_{2}(s) $$${\Rightarrow \frac{\Delta H_{2} - 24}{523 - 473} = 0.055 - 0.031 }{\therefore\Delta H_{2} = 25.2\text{ cal/gm}}$$
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