Chapter 5, Problem 5.112AP

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given reaction takes place in the Styrofoam cup is to be stated.

The amount of $\text{NaOH}$ or ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left in the Styrofoam cup is to be predicted.

The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given reaction is to be calculated.

Concept introduction: In a balanced chemical equation the number of atom in product and reactant side are equal.

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

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

The enthalpy change per mole of an aqueous solution in the reaction is calculated by the formula,

$\Delta H=\frac{m\times s\times \Delta T}{n}$

To determine: The balanced chemical equation for the given reaction takes place in the Styrofoam cup.

Answer to Problem 5.112AP

Solution

The balanced chemical equation for the given reaction is,

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

Explanation of Solution

Explanation

The reaction between $\text{NaOH}\left(aq\right)$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$ is an example of neutralization reaction and written as,

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

In the above equation the number of sodium and hydrogen atoms are not balanced

Hence, it is not a balanced chemical equation. In balanced chemical equation the number of atom in product and reactant side are equal.

To balance a general equation the given steps are followed.

• The first step is to balance the number of metals.
• The second step is to balanced the number of oxygen and hydrogen.

In the above equation the sodium atoms present on right side are twice to the sodium atoms present on left side. The sodium atom is balanced by multiplying the $\text{NaOH}$ base with two.

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

Now, the hydrogen atoms present on left side are twice to the hydrogen atoms present on right side. The hydrogen atom is balanced by multiplying the ${\text{H}}_2\text{O}$ molecule with two.

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

Now, it is balanced chemical equation.

(b)

Interpretation Introduction

To determine: The amount of $\text{NaOH}$ or ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left in the Styrofoam cup.

Answer to Problem 5.112AP

Solution

There is no amount of $\text{NaOH}$ or ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left in the Styrofoam cup

Explanation of Solution

Explanation

Given

Volume of $\text{NaOH}$ sample is $100.0\text{mL}$.

Molarity $\left(\text{M}\right)$ of $\text{NaOH}$ is $1.0\text{M}$.

Volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ sample is $50.0\text{mL}$.

Molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $1.0\text{M}$.

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,

$\text{M}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\left(n\right)}{\text{Vol}\text{.}\text{of}\text{solution}\left(V\right)\text{in}\text{mL}}\times 1000$ (1)

Substitute the values of Molarity and volume of solution of $\text{NaOH}$ in equation (1).

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

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 (1).

$\begin{array}{c}\text{1}\text{mol}/\text{mL}=\frac{n_2}{50\text{mL}}\times 1000\\ n_2=\frac{50}{1000}\text{mol}\\ n_2=\text{0}\text{.05}\text{mol}\end{array}$

Where,

• $n_2$ is number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ used for $\text{50}\text{mL}$.

The number of moles of $\text{NaOH}$ $\left(n_1\right)$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\left(n_2\right)$ are in the ratio of $2:1$. This means the two moles of $\text{NaOH}$ react with one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The number of moles is exactly same in the case of its balanced chemical equation form as explained.

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

Hence, there is no amount of $\text{NaOH}$ or ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left in the Styrofoam cup.

(c)

Interpretation Introduction

To determine: The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given reaction

Answer to Problem 5.112AP

Solution

The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{+114.114kJ/mol}$.

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 $50.0\text{mL}$.

The initial temperature $\left(T_1\right)$ of each solution before mixing is $22.3°\text{C}$.

The final temperature $\left(T_2\right)$ of mixed solution is $31.4°\text{C}$.

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 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the reaction is calculated by the formula,

$\Delta H=\frac{m_{\text{Mixed}}\times s\times \Delta T}{n_2}$ (2)

Where,

• $\Delta H$ is the change in enthalpy.
• $m$ is the mass of mixed solution.
• $s$ is the specific heat of mixed solution.
• $\Delta T$ is the change in temperature.
• $n_2$ is the number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The change in temperature $\left(\Delta T\right)$ is calculated by,

$\Delta T=T_2-T_1$ (3)

Where,

• $T_2$ is final temperature of mixed solution.
• $T_1$ is initial temperature of each solution before mixing.

Substitute the values of $T_2$ and $T_1$ in equation (3).

$\begin{array}{c}\Delta T=31.4°\text{C}-22.3°\text{C}\\ \Delta T\text{=9}\text{.1}°\text{C}\end{array}$

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

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

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 (4).

$\begin{array}{c}{V}_{\text{Mixed}}=100.0\text{mL}+50.0\text{mL}\\ \text{=150}\text{.0}\text{mL}\end{array}$

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

$m_{\text{Mixed}}=V_{\text{Mixed}}\times {\rho }_{\text{Mixed}}$ (5)

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 (5).

$\begin{array}{c}{m}_{\text{Mixed}}=150.0\text{mL}\times 1\text{g}/\text{mL}\\ \text{=}150.0\text{g}\end{array}$

Substitute the value of $m_{\text{Mixed}}$, $s$, $\Delta T$ and $n_2$ in equation (2).

$\begin{array}{c}\Delta H=\frac{150\text{g}\times 4.18\text{J}/\left(\text{g}\text{.}°\text{C}\right)\times \text{9}\text{.1}°\text{C}}{\text{0}\text{.05}\text{mol}}\\ =1,14,114\text{J}/\text{mol}\end{array}$

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

$\begin{array}{c}1\text{J}/{\text{mol=10}}^{-3}\text{kJ}/\text{mol}\\ \text{1,14,114}\text{J}/\text{mol}=\text{1,14,114}\times {\text{10}}^{-3}\text{kJ}/\text{mol}\\ =114.114\text{kJ}/\text{mol}\end{array}$

In the statement it is given that, no heat is lost during this neutralization reaction. That means the final value of $\Delta H$ would be in positive.

Hence, the enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{+114.114kJ/mol}$.

Conclusion

1. a. The balanced chemical equation for the given reaction is,

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

2. b. There is no amount of $\text{NaOH}$ or ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ left in the Styrofoam cup.
3. c. The enthalpy change per mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{+114.114kJ/mol}$