Question

The specific heats of iodine vapor and solid are \[0.031\]and $0.05{\text{cal}}.{{\text{g}}^{ - 1}}$ respectively. If heat of sublimation of iodine is $24{\text{cal}}.{{\text{g}}^{ - 1}}$ at ${200^ \circ }{\text{C}}$. Calculate its value at ${250^ \circ }{\text{C}}$.

Answer 1

Hint: Sublimation is the process where a solid changes to vapor without passing through the liquid state. Heat capacity is the amount of heat needed to increase the system’s temperature by ${1^ \circ }{\text{C}}$ or ${\text{1K}}$. It is calculated by dividing the amount of heat, ${\text{Q}}$ by change in temperature, $\Delta {\text{T}}$.

Complete step by step solution:
Specific heat capacity is the capacity of a substance to store heat depending on its chemical composition. The formula for finding specific heat capacity is given below:
Specific heat capacity, ${\text{C}} = \dfrac{{\text{Q}}}{{{\text{m}}\Delta {\text{T}}}}$, where ${\text{Q}}$ is the amount of heat, ${\text{m}}$ is the mass of the substance and $\Delta {\text{T}}$ is the change in temperature.
While molar heat capacity of a substance is the heat capacity per mole of the substance. It is denoted by ${{\text{C}}_{\text{p}}}$. In molar heat capacity, ${\text{m}}$ will be the molar mass.
It is given that:
Specific heat capacity of iodine solid, ${{\text{C}}_1} = 0.05{\text{cal}}.{{\text{g}}^{ - 1}}$
The specific heat capacity of iodine vapor, ${{\text{C}}_2}{\text{ = 0}}{\text{.031cal}}{\text{.}}{{\text{g}}^{ - 1}}$
 Initial temperature, ${{\text{T}}_1} = {200^ \circ }{\text{C}}$, Final temperature, ${{\text{T}}_2} = {250^ \circ }{\text{C}}$
 Heat of sublimation at ${{\text{T}}_1} = {200^ \circ }{\text{C}}$ $\Delta {{\text{H}}_1} = 24{\text{cal}}.{{\text{g}}^{ - 1}}$
We know that the molecular weight of iodine, ${\text{m = 127g}}$
Total molar heat capacity, ${{\text{C}}_{\text{p}}}$ is equal to the difference of molar heat capacity of products, \[{{\text{C}}_{{\text{p}}\left( {{\text{pro}}} \right)}}\] and reactants, \[{{\text{C}}_{{\text{p}}\left( {{\text{rea}}} \right)}}\].
i.e. \[{{\text{C}}_{\text{p}}} = {{\text{C}}_{{\text{p}}\left( {{\text{pro}}} \right)}} - {{\text{C}}_{{\text{p}}\left( {{\text{rea}}} \right)}} \to \left( 1 \right)\]
Molar heat capacity can be calculated by multiplying specific heat capacity with molecular weight.
i.e. molar heat capacity of products, \[{{\text{C}}_{{\text{p}}\left( {{\text{pro}}} \right)}} = {{\text{C}}_2} \times {\text{m = 0}}{\text{.031}} \times {\text{127 = 3}}{\text{.94}} \to \left( 2 \right)\]
Molar heat capacity of reactants, \[{{\text{C}}_{{\text{p}}\left( {{\text{rea}}} \right)}} = {{\text{C}}_1} \times {\text{m = 0}}{\text{.05}} \times {\text{127 = 6}}{\text{.35}} \to \left( 3 \right)\]
Substituting $\left( 2 \right),\left( 3 \right)$ in $\left( 1 \right)$, we get
\[{{\text{C}}_{\text{p}}} = {{\text{C}}_{{\text{p}}\left( {{\text{pro}}} \right)}} - {{\text{C}}_{{\text{p}}\left( {{\text{rea}}} \right)}} = 3.9 - 6.35 = - 2.45{\text{cal}}.{{\text{g}}^{ - 1}}\]
Kirchoff equation gives the variation of the heat of reaction given by the formula:
${{\text{C}}_{\text{p}}} = \dfrac{{\Delta {{\text{H}}_2} - \Delta {{\text{H}}_1}}}{{{{\text{T}}_2} - {{\text{T}}_1}}}$, $\Delta {{\text{H}}_2}$ is the heat of sublimation at ${{\text{T}}_2} = {250^ \circ }{\text{C}}$
$\Rightarrow {\text{ - 2}}{\text{.45}} = \dfrac{{\Delta {{\text{H}}_2} - 24}}{{250 - 200}} \Rightarrow - {\text{2}}{\text{.45}} = \dfrac{{\Delta {{\text{H}}_2} - 24}}{{50}}$
$\Rightarrow \Delta {{\text{H}}_2} - 24 = - 2.45 \times 50 \Rightarrow \Delta {{\text{H}}_2} - 24 = - 122.5$
Thus $\Delta {{\text{H}}_2} = - 122.5 + 24 = - 98.5{\text{cal}}.{{\text{g}}^{ - 1}}$

Note: Heat capacity of a substance is an intrinsic property. Latent heat and specificity are different. Latent heat of melting is the heat needed to melt a substance. When the substance is melting, the temperature usually remains constant till it is completely molten. Specific heat relates heat to a change in temperature.