Question

Question 85. Find the maximum possible temperature that may be reached when $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}\mathbf{mol}$role="math" localid="1663438895360" $\mathit{C}\mathit{a}\mathbf{(}\mathbf{O}\mathbf{H}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$is allowed to react with role="math" localid="1663438958928" $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{L}$of a $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{M}$$\mathbf{HCl}$ solution, both initially at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$. Assume that the final volume of the solution is$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{L}$ , and that the specific heat at constant pressure of the solution is constant and equal to that of water,$\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathit{g}}^{\mathbf{-}\mathbf{1}}$ .

Short answer

The maximum temperature of this system is 356.25 K.

Step-by-step solution

Given data

All date given in the question is written as:

$\begin{array}{c}n\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)=0.050\mathrm{mol}\\ V\left(\mathrm{HCl}\right)=1.00\mathrm{L}\\ c\left(\mathrm{HCl}\right)=1.00\mathrm{M}\\ V_{\mathrm{final}}=1.00\mathrm{L}\end{array}$

The standard enthalpies of formation $\Delta H_f\left({\mathrm{CaCl}}_2,s\right)=-795.8\mathrm{kJ}{\mathrm{mol}}^{-1}$

$\begin{array}{c}\Delta H_f\left({\mathrm{H}}_2\mathrm{O},l\right)=-285.83\mathrm{kJ}{\mathrm{mol}}^{-1}\\ \Delta H_f\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2,s\right)=-986.09\mathrm{kJ}\mathrm{mol}-1\\ \Delta H_f(\mathrm{HCl},aq)=-92.90\mathrm{kJ}\mathrm{mol}-1\end{array}$

.

Concept of change in enthalpy of reaction

Enthalpy change is the standard enthalpy of formation, which has been determined for a vast number of substances.

The reactants undergo chemical changes and combine to produce products in any general chemical reaction.

It can be represented by the following equation:$\mathbf{Reactant}\to \mathbf{Product}$

For any such reaction, the change in enthalpy is represented as$\mathit{\Delta }\mathit{H}$ and is termed as the reaction enthalpy.

The reaction enthalpy is calculated as $\mathit{\Delta }\mathit{H}\mathbf{=}\mathit{\Delta }{\mathit{H}}_{\mathbf{1}}\mathbf{+}\mathit{\Delta }{\mathit{H}}_{\mathbf{2}}$.

Where, $\Delta H_1$is change in enthalpy of reactant and $\Delta H_2$ is change in enthalpy of product.

Calculate total enthalpy change

Change in enthalpy is calculated by formula which is

$\Delta H=\sum \Delta H_f\left(products\right)-\sum \Delta H_f\left(\mathrm{reactants}\right)$

For this reaction the formula becomes $\Delta H=\Delta H_f\left({\mathrm{CaCl}}_2,s\right)+2\times \Delta H_f\left({\mathrm{H}}_2\mathrm{O},l\right)-\left(\Delta H_f\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2,s\right)+2\times \Delta H_f(\mathrm{HCl},aq)\right).$

Calculate the enthalpy changes for 1 mol of calcium hydroxide.

$\begin{array}{l}\Delta H_m=-795.8+2\times (-285.83)-(-986.09-2\times 92.90)\\ \Delta H_m=-195.57\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Calculate the enthalpy change for of calcium hydroxide.

$\begin{array}{l}\Delta H=n\times \Delta H_m\\ \Delta H=0.050\times (-195.57)\\ \Delta H=-9.78\mathrm{kJ}\end{array}$

The heat is equal to change in enthalpy with opposite sign.

So,

$q=9.78\mathrm{kJ}$

Evaluation of total mass

Calculate the mass of calcium hydroxide by given formula.

$m=n\times M$

The value of above-mentioned terms is:

data-custom-editor="chemistry" $\begin{array}{c}M\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)=74.09\mathrm{g}{\mathrm{mol}}^{-1}\\ n\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)=0.050\mathrm{mol}\\ m=0.050\times 74.09\\ m\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)=3.705\mathrm{g}\end{array}$

Calculate the mass of hydrochloric acid

$m=c\times V\times M$ …… (1)

Since, $n=c\times V$.

Therefore, $m=c\times V\times M$. ……. (2)

$\begin{array}{c}V\left(\mathrm{HCl}\right)=1.00\mathrm{L}\\ c\left(\mathrm{HCl}\right)=1.00\mathrm{M}\\ M\left(\mathrm{HCl}\right)=36.46\mathrm{g}{\mathrm{mol}}^{-1}\end{array}$

Put the value of above term in equation (2).

$\begin{array}{c}m=1.00\times 1.00\times 36.46\\ m\left(\mathrm{HCl}\right)=36.46\mathrm{g}\end{array}$

The total mass is determined by:

$\begin{array}{l}m\left(to\mathrm{tal}\right)=m\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)+m\left(\mathrm{HCl}\right)\\ m\left(total\right)=3.705+36.46\\ m\left(total\right)=40.17\mathrm{g}\end{array}$

Calculation of temperature change

Calculate the temperature change by given formula.

$\begin{array}{l}\Delta T=\frac{q}{m{c}_p}\\ \Delta T=\frac{9.78\times {10}^3}{40.17\times 4.18}\\ \Delta T=58.25\mathrm{K}\end{array}$

Calculation of maximum temperature

The maximum temperature is calculated as:

$\begin{array}{c}\Delta T=T_{\max }-T_i\\ T_{\max }=\Delta T+T_i.\\ T_{\max }=58.25+298\\ T=356.25\mathrm{K}\end{array}$