Chapter 5, Problem 5.113AP

Interpretation Introduction

Interpretation: The maximum measured temperature in the Styrofoam cup is to be predicted for the given data.

Concept introduction: The maximum temperature $\left({\text{T}}_{\text{2}}\right)$ in the Styrofoam cup is calculated by the formula,

$T_2=\frac{\Delta H_{\text{NaOH}}\times n_1}{m_{\text{Mixed}}\times s}+T_1$

To determine: The maximum measured temperature in the Styrofoam cup for the given data.

Answer to Problem 5.113AP

Solution

The maximum measured temperature in the Styrofoam cup is $\underset{\_}{33.27°C.}$

Explanation of Solution

Explanation

Given

The volume of $\text{NaOH}$ solution $\left(V_{\text{NaOH}}\right)$ is $100.0\text{mL}$.

The volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution $\left(V_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)$ is $65.0\text{mL}$.

The initial temperature $\left(T_1\right)$ of both solution before mixing is $25.0°\text{C}$.

The density $\left({\rho }_{\text{Mixed}}\right)$ of mixed solution is $1.00\text{g}/\text{mL}$.

The specific heat $\left(s\right)$ of mixed solution is $4.18\text{J}/\left(\text{g}\text{.}°\text{C}\right)$.

The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\left(\Delta H_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)$ is $+114.114\text{kJ}/\text{mol}$.

The maximum temperature $\left({\text{T}}_{\text{2}}\right)$ in the Styrofoam cup is calculated by the formula,

$\begin{array}{c}\Delta H_{\text{NaOH}}=\frac{m_{\text{Mixed}}\times s\times \Delta T}{n_1}\\ \Delta T=\frac{\Delta H_{\text{NaOH}}\times n_1}{m_{\text{Mixed}}\times s}\\ T_2-T_1=\frac{\Delta H_{\text{NaOH}}\times n_1}{m_{\text{Mixed}}\times s}\\ T_2=\frac{\Delta H_{\text{NaOH}}\times n_1}{m_{\text{Mixed}}\times s}+T_1\end{array}$ (1)

Where,

• $\Delta H_{\text{NaOH}}$ is change in enthalpy per mole of $\text{NaOH}$.
• $m$ is mass of mixed solution.
• $s$ is specific heat of mixed solution.
• $n_1$ is number of moles of $\text{NaOH}$.
• $T_2$ is maximum measured temperature.
• $T_1$ is initial temperature of both solution.

The number of moles of solute dissolved per liter of the solution is stated as Molarity.

$\text{M}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\left(n\right)}{\text{Vol}\text{.}\text{of}\text{solution}\left(V\right)\text{in}\text{Litre}}$

If the amount of solution is given in $\text{mL}$, then the formula of molarity is rewritten as,

$\text{M}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\left(n\right)}{\text{Vol}\text{.}\text{of}\text{solution}\left(V\right)\text{in}\text{mL}}\times 1000$ (2)

Substitute the values of Molarity and volume of solution of $\text{NaOH}$ in equation (2).

$\begin{array}{c}\text{1}\text{mol}/\text{mL}=\frac{n_1}{\text{100}\text{mL}}\times 1000\\ n_1=\frac{100}{1000}\text{mol}\\ n_1=\text{0}\text{.10}\text{mol}\end{array}$

Where,

• $n_1$ is number of moles of $\text{NaOH}$ used for $\text{100}\text{mL}$.

Substitute the values of Molarity and volume of solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in equation (2).

$\begin{array}{c}\text{1}\text{mol}/\text{mL}=\frac{n_2}{65\text{mL}}\times 1000\\ n_2=\frac{65}{1000}\text{mol}\\ n_2=\text{0}\text{.065}\text{mol}\end{array}$

Where,

• $n_2$ is number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ used for $\text{65}\text{.0}\text{mL}$.

The balanced chemical equation for the given reaction is,

$\text{2NaOH}\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

In the above balanced chemical equation, the number of moles of $\text{NaOH}$ is twice to number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. However, in actual the number of moles of $\text{NaOH}$ $\left(n_1\right)$ are less than twice to the number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. This means $\text{NaOH}$ acts as a limiting reactant.

The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\left(\Delta H_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)$ is $+114.114\text{kJ}/\text{mol}$. The enthalpy change per mole of $\text{NaOH}$ $\left(\Delta H_{\text{NaOH}}\right)$ is calculated by,

$\begin{array}{c}\Delta H_{\text{NaOH}}=\frac{\Delta H_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}}{2}\\ =\frac{114.114\text{kJ}/\text{mol}}{2}\\ \Delta H_{\text{NaOH}}=57.057\text{kJ}/\text{mol}\end{array}$

Conversion of $\text{kJ}/\text{mol}$ to $\text{J}/\text{mol}$.

$\begin{array}{c}1\text{kJ}/\text{mol}={\text{10}}^3\text{J}/\text{mol}\\ 57.057\text{kJ}/\text{mol}=57.057\times {\text{10}}^3\text{J}/\text{mol}\\ =57057\text{J}/\text{mol}\end{array}$

So, the value of $\Delta H_{\text{NaOH}}$ in $\text{J}/\text{mol}$ is $57057\text{J}/\text{mol}$.

The volume of mixed solution $\left(V_{\text{Mixed}}\right)$ is the sum of the volume of $\text{NaOH}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

$V_{\text{Mixed}}=V_{\text{NaOH}}+V_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ (3)

Where,

• $V_{\text{NaOH}}$ is volume of $\text{NaOH}$ solution.
• $V_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ is volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.

Substitute the value of $V_{\text{NaOH}}$ and $V_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$ in equation (3).

$\begin{array}{c}{V}_{\text{Mixed}}=100.0\text{mL}+65.0\text{mL}\\ =\text{165}\text{.0}\text{mL}\end{array}$

The mass of the mixed solution $\left(m_{\text{Mixed}}\right)$ is calculated by the formula,

$m_{\text{Mixed}}=V_{\text{Mixed}}\times {\rho }_{\text{Mixed}}$ (4)

Where,

• ${\rho }_{\text{Mixed}}$ is density of mixed solution.
• $V_{\text{Mixed}}$ is volume of mixed solution.

Substitute the value of $V_{\text{Mixed}}$ and ${\rho }_{\text{Mixed}}$ in equation (4).

$\begin{array}{c}{m}_{\text{Mixed}}=165.0\text{mL}\times 1\text{g}/\text{mL}\\ =165.0\text{g}\end{array}$

Substitute the value of $m_{\text{Mixed}}$, $s$, $T_1$ $\Delta H_{\text{NaOH}}$ and $n_1$ in equation (1).

$\begin{array}{c}{T}_2=\frac{\Delta H_{\text{NaOH}}\times n_1}{m_{\text{Mixed}}\times s}+T_1\\ T_2=\frac{57057\text{J}/\text{mol}\times \text{0}\text{.10}\text{mol}}{165.0\text{g}\times 4.18\text{J}/\left(\text{g}\text{.}°\text{C}\right)}+25.0°\text{C}\\ \text{=}\frac{5705.7\text{J}}{689.7\text{J}/°\text{C}}+25.0°\text{C}\\ T_2=\underset{\_}{33.27°C}\end{array}$

Hence, the maximum measured temperature in the Styrofoam cup is $\underset{\_}{33.27°C.}$

Conclusion

The maximum measured temperature in the Styrofoam cup is $\underset{\_}{33.27°C.}$