Chapter 5, Problem 138AP

Interpretation Introduction

Interpretation:

The change in temperature for a reaction is to be determined.

Concept introduction:

Heat absorbed or released in a reaction is associated with the change in temperature in the reaction, which is determined as follows:

$q=sm\Delta T$

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

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

The concentration (molarity) of a solution is determined by the following equation:

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

Where, $m$ is the number of moles and $V$ is the volume of solution.

Answer to Problem 138AP

Solution: The change in temperature is ${0.0906}^{\circ }\text{C}$ and the final temperature is $\approx {21.1}^{\circ }\text{C}$.

Explanation of Solution

Given information: A $\text{50 ml}$ of $\text{0}\text{.0135 M HBr}$ is mixed with $\text{50 ml}$ of $\text{0}\text{.00755 M Ba}{\left(\text{OH}\right)}_2$ in coffee-cup calorimeter at room temperature $\left({21}^{\circ }\text{C}\right)$. The molar heat of neutralization is $-56.2\text{ kJ}/\text{mol}$.

The concentration (molarity) of a solution is determined by the following equation:

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

It can be rewritten as:

$M\times V=m$

The equation for ionization of $\text{HBr}$ is as follows:

$\text{HBr}\to {\text{H}}^{+}+{\text{Br}}^{-}$

From the reaction above, it is clear that the number of moles of HBr and H are equal.

So, the number of moles of ${\text{H}}^{+}$ is calculated as follows:

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

Substitute $\text{0}\text{.0135 M}$ for $M_{{\text{H}}^{+}}$ and $\text{50 ml}$ for $V_{{\text{H}}^{+}}$:

$\begin{array}{c}{m}_{{\text{H}}^{+}}=\text{0}\text{.0135 M}\times \text{50 mL}\\ =\frac{\text{0}\text{.0135 M}\times \text{50 L}}{1000}\\ =6.75\times {10}^{-4}\text{ mol}\end{array}$

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

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

From the reaction, it is clear that the number of moles of $\text{Ba}{\left(\text{OH}\right)}_2$ and ${\text{OH}}^{-}$ are equal.

So, the number of moles of ${\text{OH}}^{-}$ can be calculated as follows:

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

Substitute $\text{0}\text{.00755 M}$ for $M_{{\text{OH}}^{-}}$ and $\text{50 ml}$ for $V_{{\text{OH}}^{-}}$:

$\begin{array}{c}{n}_{{\text{OH}}^{-}}=\text{0}\text{.00755 M}\times \text{50 mL}\times \text{2}\\ =\frac{\text{0}\text{.00755 M}\times \text{50 mL}}{1000}\\ =7.50\times {10}^{-4}\text{ mol}\end{array}$

So, $\text{HBr}$ is a limiting reagent.

Now, heat produced by the reaction is calculated as:

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

Substitute $-56.2\text{ kJ}/\text{mol}$ for $\Delta {\text{H}}^{\circ }$ and $6.75\times {10}^{-4}\text{ mol}$ for $\left(m_{\text{HBr}}\right)$:

$\begin{array}{c}{q}_{\text{rxn}}=6.75\times {10}^{-4}\text{ mol}\times -56.2\text{ kJ}/\text{mol}\\ \text{=}-379.3\times {10}^{-4}\text{ kJ}\\ \text{=}-379.3\times {10}^{-4}\times {10}^3\text{ kJ}\\ \text{=}-37.9\text{ J}\end{array}$

As it is known that $q_{\text{rxn}}=-q_{\text{surr}}$, therefore $q_{\text{surr}}=37.9\text{ J}$.

So, heat is given by:

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

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

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

Here, $m_{{}_{{\text{H}}_2\text{O}}}$ is the mass of ${\text{H}}_2\text{O}$, $V_{{}_{{\text{H}}_2\text{O}}}$ is the volume of ${\text{H}}_2\text{O}$, and $d_{{}_{{\text{H}}_2\text{O}}}$ is the density of ${\text{H}}_2\text{O}$.

Substitute $100\text{ mL}$ for $V_{{}_{{\text{H}}_2\text{O}}}$ and $\text{1 g/mL}$ for $d_{{}_{{\text{H}}_2\text{O}}}$:

$\begin{array}{c}{m}_{{}_{{\text{H}}_2\text{O}}}=100\text{ mL}\times \text{1 g/mL}\\ =100\text{ g}\end{array}$

Now, substitute $-37.9\text{ J}$ for $q_{\text{surr}}$, $\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}$ for $s_{{\text{H}}_2\text{O}}$ and $\text{100 g}$ for $m_{{}_{{\text{H}}_2\text{O}}}$ in equation $\left(1\right)$:

$\begin{array}{c}\text{37}\text{.9 J}=\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}\times \text{100 g}\times \Delta T\\ \text{37}\text{.9 J}=418.4\text{ J}{/}^{\text{o}}\text{C}\times \Delta T\\ \frac{\text{37}\text{.9 J}}{418.4\text{ J}{/}^{\text{o}}\text{C}}=\Delta T\\ {0.0906}^{\text{o}}\text{C}=\Delta T\end{array}$

Now, the change in temperature is determined by the following equation:

$\Delta T=\left(T_f-T_i\right)$…… (2)

Here, $\Delta T$ is the change in temperature, $T_f$ is the final temperature, and $T_i$ is the initial temperature.

Equation $\left(2\right)$ can be rewritten as follows:

$T_f=\left(\Delta T+T_i\right)$

Substitute ${0.0906}^{\text{o}}\text{C}$ for $\Delta T$ and ${21}^{\circ }\text{C}$ for $T_i$:

$\begin{array}{c}{T}_f=\left({0.0906}^{\text{o}}\text{C}+{21}^{\circ }\text{C}\right)\\ ={21.0906}^{\text{o}}\text{C}\\ \approx \text{21}{\text{.1}}^{\text{o}}\text{C}\end{array}$

Conclusion

The change in temperature for the given reaction is ${0.0906}^{\circ }\text{C}$ and the final temperature achieved is approximately equal to ${21.1}^{\circ }\text{C}$.