Chapter 6, Problem 6.94P

(a)

Interpretation Introduction

Interpretation:

The mass of $\text{AgI}$ formed when ${\text{AgNO}}_3$ reacted with $\text{NaI}$ is to be calculated.

Concept introduction:

A limiting reagent is the species that is completely consumed in a chemical reaction. The amount of product formed or reactant reacted in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Answer to Problem 6.94P

The mass of $\text{AgI}$ formed is $\text{0}\text{.35}\text{g}$.

Explanation of Solution

The equation of the chemical reaction between ${\text{AgNO}}_3$ and $\text{NaI}$ is,

${\text{AgNO}}_3\left(aq\right)+\text{NaI}\left(aq\right)\to \text{AgI}\left(s\right)+{\text{NaNO}}_3\left(aq\right)$

The formula to calculate moles of ${\text{AgNO}}_3$ is,

${\text{Moles of AgNO}}_3=\left({\text{volume of AgNO}}_3\right)\left(\frac{{\text{ mass of AgNO}}_3}{{\text{molar mass of AgNO}}_3}\right)$ (1)

Substitute $50.0\text{mL}$ for volume of ${\text{AgNO}}_3$, $5.0\text{g/L}$ for mass of ${\text{AgNO}}_3$ and $169.9\text{g/mol}$ for molar mass of ${\text{AgNO}}_3$ in the equation(1).

$\begin{array}{c}{\text{Moles of AgNO}}_3=\left(50.0\text{mL}\right)\left(\frac{5.0\text{g/L}}{169.9\text{g/mol}}\right)\\ =1.47145\times {10}^{-3}\text{mol}\end{array}$

The formula to calculate moles of $\text{NaI}$ is,

$\text{Moles of NaI}=\left(\text{volume of NaI}\right)\left(\frac{\text{ mass of NaI}}{\text{molar mass of NaI}}\right)$ (2)

Substitute $50.0\text{mL}$ for volume of $\text{NaI}$, $5.0\text{g/L}$ for mass of $\text{NaI}$ and $149.9\text{g/mol}$ for molar mass of $\text{NaI}$ in the equation(2).

$\begin{array}{c}\text{Moles of NaI}=\left(50.0\text{mL}\right)\left(\frac{5.0\text{g/L}}{149.9\text{g/mol}}\right)\\ =1.677785\times {10}^{-3}\text{mol}\end{array}$

Moles of ${\text{AgNO}}_3$ is less so ${\text{AgNO}}_3$ is the imiting reagent.

The formula to calculate mass of $\text{AgI}$ is,

$\text{Mass of AgI}=\left({\text{moles of AgNO}}_3\right)\left(\frac{\text{1}\text{mol}\text{AgI}}{\text{1}\text{mol}{\text{AgNO}}_3}\right)\left(\text{molar mass of AgI}\right)$ (3)

Substitute $1.47145\times {10}^{-3}\text{mol}$ for moles of ${\text{AgNO}}_3$ and $234.8\text{g/mol}$ for molar mass of $\text{AgI}$ in the equation (3).

$\begin{array}{c}\text{Mass of AgI}=\left(1.47145\times {10}^{-3}\text{mol}\right)\left(\frac{\text{1}\text{mol}\text{AgI}}{\text{1}\text{mol}{\text{AgNO}}_3}\right)\left(234.8\text{g/mol}\right)\\ =0.345496\text{g}\\ \approx \text{0}\text{.35}\text{g}\end{array}$

Conclusion

The mass of $\text{AgI}$ formed is $\text{0}\text{.35}\text{g}$.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ of the following reaction is to be determined.

${\text{Ag}}^{+}\left(aq\right)+{\mathrm{I}}^{-}\left(aq\right)\to \text{Ag}\mathrm{I}\left(s\right)$

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.94P

$\Delta H_{\text{rxn}}^{°}$ of the following reaction is $–112.3\text{kJ}$.

Explanation of Solution

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\left\{1\Delta H_{\text{f}}^{°}\left[\text{Ag}\mathrm{I}\left(s\right)\right]\right\}-\left\{1\Delta H_{\text{f}}^{°}\left[{\text{Ag}}^{+}\left(aq\right)\right]+1\Delta H_{\text{f}}^{°}\left[{\mathrm{I}}^{-}\left(aq\right)\right]\right\}\right]$ (4)

Substitute $-62.38\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{Ag}\mathrm{I}\left(s\right)\right]$, $-105.9\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{Ag}}^{+}\left(aq\right)\right]$ and $-55.94\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\mathrm{I}}^{-}\left(aq\right)\right]$ in the equation (4).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(1\text{mol}\right)\left(-62.38\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-105.9\text{kJ/mol}\right)+\left(1\text{mol}\right)\left(-55.94\text{kJ/mol}\right)\right\}\end{array}\right]\\ =–112.34\text{kJ}\\ \approx –112.3\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{°}$ of the following reaction is $–112.3\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

$\Delta T_{\text{sol}}$ of $\text{AgI}$ is to determined

Concept introduction:

Heat capacity $\left(C\right)$ of a material is the quantity to measure the heat needed to raise the temperature of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

$q=\left(C\right)\left(\Delta T\right)$ (5)

Here,

$\Delta T$ is the temperature difference.

$q$ is the heat released or absorbed.

$C$ is the heat capacity of the substance.

Specific heat capacity $\left(c\right)$ of a substance is the quantity to measure the heat needed to raise the temperature of $1\text{g}$ of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

$q=\left(\text{mass}\right)\left(c\right)\left(\Delta T\right)$ (6)

Here,

$\Delta T$ is the temperature difference.

$q$ is the heat released or absorbed.

$c$ is the specific heat capacity of the substance.

Answer to Problem 6.94P

$\Delta T_{\text{sol}}$ is $\text{0}\text{.40}\text{K}$.

Explanation of Solution

The equation of the chemical reaction between ${\text{AgNO}}_3$ and $\text{NaI}$ is,

${\text{AgNO}}_3\left(aq\right)+\text{NaI}\left(aq\right)\to \text{AgI}\left(s\right)+{\text{NaNO}}_3\left(aq\right)$

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ is,

$\Delta H_{\text{rxn}}^{°}=\left(\Delta H_{\text{rxn}}^{°}\text{of AgI}\right)\left(\frac{1\text{mol}\text{AgI}}{1\text{mol}{\text{AgNO}}_3}\right)\left({\text{moles of AgNO}}_3\right)$ (7)

Substitute $112.34\text{kJ}$ for $\Delta H_{\text{rxn}}^{°}$ of $\text{AgI}$ and $1.47145\times {10}^{-3}\text{mol}$ for moles of ${\text{AgNO}}_3$ in the equation (7).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left(112.34\text{kJ}\right)\left(\frac{1\text{mol}\text{AgI}}{1\text{mol}{\text{AgNO}}_3}\right)\left(1.47145\times {10}^{-3}\text{mol}\right)\\ =165.3026\times {10}^{-3}\text{kJ}\end{array}$

The formula to calculate mass of solution is,

$\text{Mass of solution}=\left(\text{density of solution}\right)\left({\text{volume of AgNO}}_3+\text{volume}of\text{NaI}\right)$ (8)

Substitute $1\text{g/mL}$ for density of solution , $50\text{mL}$ for volume of ${\text{AgNO}}_3$ and $50\text{mL}$ for volume of $\text{NaI}$ in the equation (8).

$\begin{array}{c}\text{Mass of solution}=\left(1\text{g/mL}\right)\left(50\text{mL}+50\text{mL}\right)\\ =100\text{g}\end{array}$

Rearrange the equation (6) to calculate $\Delta T$.

$\Delta T=\frac{q}{\left(\text{mass}\right)\left(c\right)}$ (9)

Substitute $165.3026\times {10}^{-3}\text{kJ}$ for $q$, $4.184\text{J/g}\cdot \text{K}$ for $c$ and $100\text{g}$ for mass in the equation (9).

$\begin{array}{c}\Delta T=\frac{165.3026\times {10}^{-3}\text{kJ}}{\left(100\text{g}\right)\left(4.184\text{J/g}\cdot \text{K}\right)}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\\ =0.39508\text{K}\\ \approx \text{0}\text{.40}\text{K}\end{array}$

Conclusion

$\Delta T_{\text{sol}}$ is $\text{0}\text{.40}\text{K}$.