Chapter 6, Problem 6.80P

You want to determine $\text{Δ}{\text{H}}^o$ for the reaction $\text{Z}n(s)+2\text{H}\text{C}\text{l}(aq)\to \text{Z}n\text{C}{\text{l}}_2(aq)+{\text{H}}_2(g)$

1. To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose $\text{Δ}\text{H}$ is known:

$\text{N}\text{a}\text{O}\text{H}(aq)+\text{H}\text{C}\text{l}(aq)\to \text{N}\text{a}\text{C}\text{l}(aq)+{\text{H}}_2\text{O}(\text{l})$ $\text{Δ}{\text{H}}^o=-57.32\text{kJ}$ Calculate the heat capacity of the calorimeter from these data:

Amounts used: 50. mL of 2.00 M HCl and

50. mL of 2.00 M NaOH

Initial T of both solutions: 16.9°C

Maximm T recorded during reaction: 30.4°C

Density of resulting $\text{NaCl solution: 1}\text{.04 g/mL}$ $\text{c of 1}\text{.00 M NaCl(aq) = 3}\text{.93 J/g}\text{.K}$

1. Use the result from part (a) and the following data to determine $\text{Δ}{\text{H}}_{rxn}^o$ for the reaction between zinc and HCl(aq):

Amount used: $\text{100}\text{. mL of 1}\text{.00 M HCl and 1}\text{.3078 g of Zn}$ Initial T of HCl solution and Zn: 16.8°C

Maximum T recorded during reaction: 24.1°C

Denstity of $1.0\text{M}\text{H}\text{C}\text{l}so\text{l}ution=1.015\text{g/mL}$ c of resulting $\text{Z}n\text{C}{\text{l}}_2(aq)=3.95\text{ J/g}\text{.K}$     

1. Given the values below, what is the error in your experiment?

$\text{Δ}{\text{H}}_f^{o}of\text{ HCl(aq) = -1}\text{.652}\times {\text{10}}^{\text{2}}\text{kJ/mol}$ $\text{Δ}{\text{H}}_f^{o}of{\text{ ZnCl}}_{\text{2}}\text{(aq) = -4}\text{.822}\times {\text{10}}^{\text{2}}\text{kJ/mol}$

Interpretation Introduction

Interpretation: The value of $\Delta H°$ for the given equation needs to be determined.

  $\text{Zn}\left(\text{s}\right)\text{+2HCl}\left(\text{aq}\right)\to {\text{ZnCl}}_{\text{2}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction: Following expression is used to determine heat generated.

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=Heat}\text{of}\text{solution+heat}\text{of}\text{calorimeter}\\ \text{=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\end{array}$

Answer to Problem 6.80P

Heat capacity of calorimeter is $15.87\frac{\text{J}}{°\text{C}}$ .

Explanation of Solution

Calculate the heat capacity of calorimeter using the following equation.

  $\text{NaOH}\left(\text{aq}\right)\text{+2HCl}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)\Delta H°=-57.32\text{kJ}$

The amounts used are as follows:

Volume of $\text{HCl}$ is 50.0mL and $\text{NaOH}$ is 50.0mL.

Molarity of $\text{HCl}$ is 2.0M and $\text{NaOH}$ is 2M.

Initial temperature is $16.9°\text{C}$ and recorded temperature is $30.4°\text{C}$ for both solutions.

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=Heat}\text{of}\text{solution+heat}\text{of}\text{calorimeter}\\ \text{=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\end{array}$

Calculate mass of $\text{NaCl}$ as follows:

At first calculate volume of $\text{NaCl}$ .

  $\begin{array}{c}\text{Volume}\text{of}\text{NaCl=Volume}\text{of}\text{NaOH+Volume}\text{of}\text{HCl}\\ \text{=50}\text{mL+50}\text{mL}\\ \text{=100}\text{mL}\end{array}$

Calculate mass of $\text{NaCl}$ as follows:

  $\begin{array}{c}\text{Mass}\text{of}\text{NaCl=Density}\text{of}\text{NaOH}\times \text{Volume}\text{of}\text{NaCl}\\ \text{=1}\text{.04}\frac{\text{g}}{\text{mL}}\times \text{50}\text{mL}\\ \text{=104}\text{g}\end{array}$

  $\begin{array}{c}\text{Moles}\text{reacted}\left(\text{HCl}\text{orNaOH}\right)\text{=Molarity}\times \text{Volume}\\ \text{=2}\frac{\text{mol}}{\text{L}}\times \text{50}\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\\ \text{=0}\text{.1}\text{mol}\end{array}$

Calculate heat of reaction as follows:

  $\begin{array}{c}\text{Heat}\text{of}\text{reaction}\text{=}\frac{-57.32\text{kJ}}{1\text{mol}}\times 0.1\text{mol}\\ \text{=}-\text{5}\text{.732}\text{kJ}\end{array}$

Therefore, heat generated by the reaction is $\text{5}\text{.732}\text{kJ}$ .

Calculate heat capacity of calorimeter is as follows:

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=Heat}\text{of}\text{solution+heat}\text{of}\text{calorimeter}\\ \text{=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\\ 5.732\text{kJ}\times \frac{1000\text{J}}{1\text{kJ}}=104\text{g}\times 3.93\frac{\text{J}}{g.\text{K}}\times \left(30.4-16.9\right)°\text{C+}{C}_{\text{calorimeter}}\left(30.4-16.9\right)°\text{C}\\ 5732\text{J=5517}\text{.72}\text{J+}{C}_{\text{calorimeter}}13.5°\text{C}\end{array}$

  $\begin{array}{c}{C}_{\text{calorimeter}}=\frac{214.28\text{J}}{13.5°\text{C}}\\ =15.87\frac{\text{J}}{°\text{C}}\end{array}$

Therefore, heat capacity of calorimeter is $15.87\frac{\text{J}}{°\text{C}}$ .

Interpretation Introduction

Interpretation: The value of $\Delta H{°}_{\text{rxn}}$ for reaction between Zinc and ${\text{ZnCl}}_2$ needs to be determined.

Concept Introduction: Following expression is used to determine heat generated.

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=Heat}\text{of}\text{solution+heat}\text{of}\text{calorimeter}\\ \text{=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\end{array}$

Answer to Problem 6.80P

Heat of reaction is $-\text{154}\text{.015}\text{kJ/mol}$ .

Explanation of Solution

Calculate heat of reaction of the following reaction:

  $\text{Zn}\left(\text{s}\right)\text{+2HCl}\left(\text{aq}\right)\to {\text{ZnCl}}_{\text{2}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)$

Initial temperature and recorded temperature is $16.8°\text{C}$ and $24.1°\text{C}$ respectively for both solutions.

Heat capacity of ${\text{ZnCl}}_{\text{2}}$ is $3.95\frac{\text{J}}{\text{g}\text{.K}}$ .

Density of $\text{HCl}$ is $1.015\frac{\text{g}}{\text{mL}}$

Calculate mass of $\text{HCl}$ as follows:

  $\begin{array}{c}\text{Mass}\text{of}\text{HCl=Density}\text{of}\text{HCl}\times \text{Volume}\text{of}\text{NaCl}\\ \text{=1}\text{.015}\frac{\text{g}}{\text{mL}}\times 10\text{0}\text{mL}\\ \text{=101}\text{.5}\text{g}\end{array}$

Calculate mass of ${\text{ZnCl}}_{\text{2}}$ as follows:

  $\begin{array}{c}\text{Mass}\text{of}{\text{ZnCl}}_2\text{=Mass}\text{of}\text{HCl+mass}\text{of}\text{Zn}\\ \text{=101}\text{.5}\text{g}+1.3078\text{g}\\ \text{=102}\text{.80}\text{g}\end{array}$

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\\ \Delta H°=102.80\text{g}\times 3.93\frac{\text{J}}{g.\text{K}}\times \left(24.1-16.8\right)°\text{C+15}\text{.87}\frac{\text{J}}{°\text{C}}\left(24.1-16.8\right)°\text{C}\\ 5732\text{J=2964}\text{.46}\text{J}+115.85\text{J}\\ =3080.31\text{J}\end{array}$

  $\begin{array}{c}\text{Moles}\text{of}\text{Zn=}\frac{\text{Mass}}{\text{Molecular}\text{weight}}\\ \text{=}\frac{1.3078\text{g}}{65.38\frac{\text{g}}{\text{mol}}}\\ \text{=0}\text{.02}\text{mol}\end{array}$

Calculate heat of reaction for 0.02 mol as follows:

  $\begin{array}{c}\Delta H°\text{=}\frac{-3080.31\text{J}}{0.02\text{mol}}\\ \text{=}-\text{154015}\text{.5}\text{J/mol}\times \frac{1\text{kJ}}{1000\text{J}}\\ \text{=}-\text{154}\text{.015}\text{kJ/mol}\end{array}$

Therefore, heat of reaction is $-\text{154}\text{.015}\text{kJ/mol}$ .

Interpretation Introduction

Interpretation: The error in the experiment needs to be determined.

Concept Introduction: Following expression is used to determine heat generated.

  $\begin{array}{c}\text{Heat}\text{generated}\text{by}\text{the}\text{reaction=Heat}\text{of}\text{solution+heat}\text{of}\text{calorimeter}\\ \text{=}{m}_{\text{solution}}{C}_{\text{solution}}\Delta T+C_{\text{calorimeter}}\Delta T\end{array}$

Answer to Problem 6.80P

Error is $2.215\frac{\text{kJ}}{\text{mol}}$ .

Explanation of Solution

Values for heat of reaction for $\text{HCl}$ and ${\text{ZnCl}}_2$ are as follows:

  $\begin{array}{c}\Delta H°={\left({{\displaystyle \sum n_i\Delta H°}}_{\text{f}}\right)}_{\text{products}}-{\left({{\displaystyle \sum n_i\Delta H°}}_{\text{f}}\right)}_{\text{reactants}}\\ =1\times {\left(\Delta H{°}_{\text{f}}\right)}_{{\text{ZnCl}}_{\text{2}}}+1\times {\left(\Delta H{°}_{\text{f}}\right)}_{{\text{H}}_{\text{2}}}-2\times 1\times {\left(\Delta H{°}_{\text{f}}\right)}_{HCl}\\ =-4.822\times {10}^2\text{kJ}-2\left(-1.652\times {10}^2\text{kJ}\right)\\ =-151.8\frac{\text{kJ}}{\text{mol}}\end{array}$

Calculate error as follows:

  $\begin{array}{c}\text{Error=True}\text{value}-\text{experiment}\text{value}\\ \text{=}-\text{151}\frac{\text{kJ}}{\text{mol}}-\left(-154.015\frac{\text{kJ}}{\text{mol}}\right)\\ =2.215\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence, error is $2.215\frac{\text{kJ}}{\text{mol}}$ .