Chapter 5, Problem 115AP

A quantity of $\text{50}\text{.0 mL of 0}\text{.200 }M\text{ Ba}{\left(\text{OH}\right)}_{\text{2}}$ is mixed with $\text{50}\text{.0 mL of 0}\text{.400 }M{\text{ HNO}}_{\text{3}}$ in a constant-pressure calorimeter having a heat capacity of $\text{496 J/C}\text{.}$ The initial temperature of both solutions is the same at $\text{22}\text{.4}°\text{C}\text{.}$ What is the final temperature of the mixed solution? Assume that the specific heat of the solutions is the same as that of water and the molar heat of neutralization is $\text{-56}\text{.2 kJ/mol}\text{.}$

Interpretation Introduction

Interpretation:

The final temperature of the reaction taking place in constant-pressure calorimeter, when given amount of $\text{0}\text{.200 M}$ $\text{Ba}{\left(\text{OH}\right)}_2$ and $\text{0}\text{.400 M}$ ${\text{HNO}}_3$ mixed, is to be calculated.

Concept introduction:

Heat absorbed or released in the reaction is given by:

$q=sm\Delta T$

Here, $s$ is the specific heat, $m$ is the mass of the sample undergoing the temperature change, and $\Delta T$ is the change in temperature.

Specific heat is the heat required to increase the temperature of $1\text{ g}$ of a substance by1°C. Its S.I. unit is $\left(\text{J}/\text{g}{\text{.}}^{\text{o}}\text{C}\right)$.

Heat absorbed or released in the reaction is also given by:

$q=C\Delta T$

Here, $C$ is the heat capacity of a body and $\Delta T$ is the change in temperature.

Heat capacity is defined as the heat required to increase the temperature of a body by1°C.

Its S.I. unit is $\left(\text{J}{/}^{\text{o}}\text{C}\right)$.

Concentration (molarity) of a solution is given by:

$M=\frac{n}{V}$

Here, $n$ is the number of moles and $V$ is the volume of the solution.

Answer to Problem 115AP

Solution: ${23.6}^{\text{o}}\text{C}$.

Explanation of Solution

Given information:

Molar heat of neutralisation is $-56.2\text{ kJ}/\text{mol}$.

$T_i={22.4}^{\circ }\text{C}$

Heat capacity $C=496\text{ J}{/}^{\circ }\text{C}$

Concentration (molarity) of a solution is given by:

$M=\frac{n}{V}$

It can be rewritten as

$M\times V=n$

Equation for the ionization of $\text{Ba}{\left(\text{OH}\right)}_2$:

$\text{Ba}{\left(\text{OH}\right)}_2\to {\text{Ba}}^{2+}+2{\text{OH}}^{-}$

So, it is clear that $M_{\text{Ba}{\left(\text{OH}\right)}_2}=M_{{\text{OH}}^{-}}$.

The number of moles of ${\text{OH}}^{-}$ can be calculated as

$M_{{\text{OH}}^{-}}\times V_{{\text{OH}}^{-}}=n_{{\text{OH}}^{-}}$

Substitute $\text{0}\text{.200 M}$ for $M_{{\text{OH}}^{-}}$ and $\text{50 ml}$ for $V_{{\text{OH}}^{-}}$

$\begin{array}{c}{n}_{{\text{OH}}^{-}}=\text{0}\text{.200 M}\times \text{50 mL}\times \text{2}\\ =\frac{\text{0}\text{.200 M}\times \text{50 mL}}{1000}\\ =0.0200\text{ mol}\end{array}$

Equation for the ionization of ${\text{HNO}}_3$:

${\text{HNO}}_3\to {\text{H}}^{+}+{\text{NO}}_3^{-}$

So, it is clear that $M_{{\text{HNO}}_3}=M_{{\text{H}}^{+}}$.

The number of moles of ${\text{H}}^{+}$ can be calculated as

$M_{{\text{H}}^{+}}\times V_{{\text{H}}^{+}}=n_{{\text{H}}^{+}}$

Substitute $\text{0}\text{.400 M}$ for $M_{{\text{H}}^{+}}$ and $\text{50 mL}$ for $V_{{\text{H}}^{+}}$

$\begin{array}{c}{n}_{{\text{H}}^{+}}=\text{0}\text{.400 M}\times \text{50 mL}\\ =\frac{\text{0}\text{.400 M}\times \text{50 mL}}{1000}\\ =0.0200\text{ mol}\end{array}$

Now, the heat produced by the reaction is calculated by:

$q=n\times \Delta {\text{H}}^{\circ }$

Substitute $-56.2\text{ kJ}/\text{mol}$ for $\Delta {\text{H}}^{\circ }$ and $0.0200\text{ mol}$ for $n$ in the above equation

$\begin{array}{c}q=0.0200\text{ mol}\times -56.2\text{ kJ}/\text{mol}\\ \text{=}-\text{1}\text{.124 kJ}\end{array}$

It is known that $q_{\text{rxn}}=-q_{\text{surr}}$.

Therefore, $q_{\text{surr}}=\text{1}\text{.124 kJ}$.

However, the heat of the surroundings includes the heat of water and the heat of the calorimeter. So,

$q_{\text{surr}}=s_{{\text{H}}_2\text{O}}{m}_{{}_{{\text{H}}_2\text{O}}}\Delta T+C_{\text{cal}}\Delta T$ …… (1)

Mass of ${\text{H}}_2\text{O}$ is calculated by:

$m_{{}_{{\text{H}}_2\text{O}}}=V_{{}_{{\text{H}}_2\text{O}}}\times d_{{}_{{\text{H}}_2\text{O}}}+n_{{}_{{\text{H}}_2\text{O}}}\times M_{{}_{{\text{H}}_2\text{O}}}$

Substitute $100\text{ mL}$ for $V_{{}_{{\text{H}}_2\text{O}}}$, $\text{1 g/mL}$ for $d_{{}_{{\text{H}}_2\text{O}}}$, $0.0200\text{ mol}$ for $n_{{}_{{\text{H}}_2\text{O}}}$, and $\text{18}\text{.016 g}$ for $M_{{}_{{\text{H}}_2\text{O}}}$ in the above equation

$\begin{array}{c}{m}_{{}_{{\text{H}}_2\text{O}}}=100\text{ mL}\times \text{1 g/mL}+0.0200\text{ mol}\times \text{18}\text{.016 g}\\ =100\text{ g}+0.36\text{ g}\\ \approx \text{100}\text{.4 g}\end{array}$

Substitute $\text{1}\text{.124 kJ}$ for $q_{\text{surr}}$, $\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}$ for $s_{{\text{H}}_2\text{O}}$, $\text{100}\text{.4 g}$ for $m_{{}_{{\text{H}}_2\text{O}}}$, $496\text{ J}{/}^{\circ }\text{C}$ for $C_{\text{cal}}$, and ${22.4}^{\circ }\text{C}$ for $T_i$ in equation (1)

$\begin{array}{c}\text{1}\text{.124 kJ }=\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}\times \text{100}\text{.4 g}\times \Delta T+{\text{ 496 J/}}^{\circ }C\times \Delta T\\ \text{1}\text{.124}\times {\text{10}}^3\text{ J=}\left[\text{420}\text{.07 J}{/}^{\circ }\text{C}+496\text{ J}{/}^{\circ }\text{C}\right]\times \Delta T\\ \text{1124}\text{.0 J =916}\text{.07 J}{/}^{\circ }\text{C}\times \Delta T\end{array}$

So, the change in temperature iscalculated as

$\begin{array}{c}\Delta T=\frac{\text{1124}\text{.0 J}}{\text{916}\text{.07 J}{/}^{\circ }\text{C}}\\ ={1.23}^{\circ }\text{C}\end{array}$

Therefore, the final temperature iscalculated as

$\begin{array}{l}\Delta T=\left(T_f-T_i\right)\\ {1.23}^{\circ }\text{C}=T_f-{22.4}^{\circ }\text{C}\\ {1.23}^{\circ }\text{C}+{22.4}^{\circ }\text{C=}{T}_f\\ {23.6}^{\circ }\text{C}=T_f\end{array}$

Conclusion

The final temperature of the reactiontaking place in constant-pressure calorimeter is ${23.6}^{\circ }\text{C}$.