Question

For ${{Ag}}$, ${{{C}}_{{p}}}\left( {{{J}}.{{K}}.{{mo}}{{{l}}^{ - 1}}} \right)$ is given by $24 + 0.006{{T}}$. If temperature of ${{3mol}}$ of silver is raised from ${{2}}{{{7}}^ \circ }{{C}}$ to its melting point ${{92}}{{{7}}^ \circ }{{C}}$ under ${{1atm}}$ pressure then $\Delta {{H}}$ is equal to:
A. ${{52kJ}}$
B. ${{76}}{{.95kJ}}$
C. ${{89}}{{.62kJ}}$
D. ${{38}}{{.62kJ}}$

Answer 1

Hint: We have to find the enthalpy change in the given problem. Enthalpy is the combination of kinetic and potential energy in a system. It is generally represented as enthalpy change which is the difference in enthalpy of reactants and products.

Complete step by step solution:
Enthalpy changes with respect to the temperature. When it changes from ${{{T}}_1}$ to ${{{T}}_2}$, it also changes from ${{{H}}_1}$ to ${{{H}}_1}$. It can be simply represented by relating the quantity of substance, the amount of heat, heat capacity, temperature change and the number of moles.
${{q}} = {{n}}{{{C}}_{{p}}}\Delta {{T}}$, where ${{q}}$ is the amount of heat, ${{n}}$ is the number of moles, ${{{C}}_{{p}}}$ is the heat capacity, $\Delta {{T}}$ is the temperature change.
It is given that heat capacity, ${{{C}}_{{p}}} = 24 + 0.006{{T}}$
Initial and final temperature, ${{{T}}_1} = 300{{K,}}{{{T}}_2} = 1200{{K}}$
The number of moles, ${{n = 3}}$
The enthalpy change can also be represented in a derivative form, i.e. ${{d}}{{{H}}_{{p}}} = {{{C}}_{{p}}}d{{T}}$
Integrating on both sides, we get
$\int_{{\operatorname{H} _1}}^{{\operatorname{H} _2}} {d{{{H}}_{{p}}}} = \int_{{{{T}}_1}}^{{{{T}}_2}} {{{{C}}_{{p}}}} d{{T}}$
When the pressure is constant, integrating ${{{C}}_{{p}}}$ with respect to the temperature, it gives the enthalpy change corresponding to the temperature change in a single phase.
$\Delta {{{H}}_{{p}}} = {{n}}\int_{{{{T}}_1}}^{{{{T}}_2}} {{{{C}}_{{p}}}} d{{T}}$
Substituting the values, we get
$\Delta {{{H}}_{{p}}} = 3 \times \int_{300}^{1200} {\left( {{{24 + 0}}{{.006T}}} \right)} d{{T}}$
${{{C}}_{{p}}}$ is in the form $a + b{{T}}$, where $a,b$ are constants and ${{T}}$ is temperature. So when it is integrated, the equation will be $a\left( {{{{T}}_2} - {{{T}}_1}} \right) + \dfrac{b}{2}\left( {{{T}}_2^2 - {{T}}_1^2} \right)$.
Here, $a = 24,b = 0.006,{{{T}}_2} = 1200{{K}},{{{T}}_1} = 300{{K}}$
So, on substituting these values in the equation we get
$\Delta {{{H}}_{{p}}} = 3\left[ {24\left( {1200 - 300} \right) + \dfrac{{0.006}}{2}\left( {{{1200}^2} - {{300}^2}} \right)} \right]$
On simplification, we get
$\Delta {{{H}}_{{p}}} = 3\left[ {24 \times 900 + 0.003\left( {1350000} \right)} \right] = 3\left( {21600 + 4050} \right) = 76950{{J}}$

Thus, the enthalpy change, $\Delta {{{H}}_{{p}}} = 76.950{{kJ}} \sim {{77kJ}}$

Note: This enthalpy change can be termed as sensible heat.
${{d}}{{{H}}_{{p}}} = {{{C}}_{{p}}}d{{T + V}}d{{P}}$ is an equation used for solids and liquids. The heat capacity of solid or liquid is not much dependent on temperature. Thus we can determine the value of enthalpy change easily as in given equation below:
$\Delta {\text{H}} = {{\text{C}}_{\text{p}}}\Delta {\text{T}}$