Chapter 4, Problem 67QRT

(a)

Interpretation Introduction

Interpretation:

The transferred heat has to be calculated.

Concept Introduction:

Specific heat can be defined as quantity of heat required to raise the temperature of $\text{1}\text{g}$ substance by $\text{1°C}$.  The relationship between heat and change in temperature can be expressed by the equation given below.

    $\text{q}=\text{cmΔT}$

Where $\text{q}$ is the Heat added

    C is the Specific heat

  $\text{m}$ is the Mass

  $\text{ΔT}$ is the Change in temperature

Answer to Problem 67QRT

The heat transferred value is $\text{2}\text{.23 kJ}\text{.}$

Explanation of Solution

Mixing of two given solutions with change in temperature, the heat transferred is determined as follows,

    $\begin{array}{c}{q}_{\text{water solution}}{\text{ = m}}_{\text{water solution}}{\text{ c}}_{\text{water solution}}\Delta {\text{T}}_{\text{water solution}}\\ \\ \Delta {\text{T}}_{\text{water solution}}\text{= 26}{\text{.18}}^{\circ }\text{C}-{24.40}^{\circ }\text{C}\\ \\ ={1.78}^{\circ }\text{C}\end{array}$

The solution mass obtained by adding the two volumes is calculated as follows,

    $\begin{array}{c}{m}_{\text{solution}}\text{ = }\left(200\text{ mL+ 100 mL}\right)\times \frac{1{\text{ g H}}_{\text{2}}\text{O}}{1{\text{ mL H}}_{\text{2}}\text{O}}\\ \\ =\text{ 300 g}\end{array}$

The heat transferred is calculated as shown below,

    $\begin{array}{c}{q}_{\text{water solution}}{\text{ = m}}_{\text{water solution}}{\text{ c}}_{\text{water solution}}\Delta {\text{T}}_{\text{water solution}}\\ \\ =\text{ 300 g }\times \left(4.184{\text{ J g}}^{\text{-1}}{}^{\circ }{\text{C}}^{\text{-1}}\right)\times \left(1.78{}^{\circ }\text{C}\right)\\ \\ =2.23\times {10}^3\text{ J}\\ \\ \text{= 2}\text{.23 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The thermochemical expression has to be identified.

Concept Introduction:

The balanced chemical equation that also defines the enthalpy change is called thermochemical equation.

Answer to Problem 67QRT

The thermochemical expression is $RbOH\left(aq\right)+HBr\left(aq\right)\to RbBr\left(aq\right)+H_{2}O\left(l\right){\Delta }_r{H}^{\circ }\text{ = -55}\text{.9 kJ}$.

Explanation of Solution

Addition of given acid and base will result in neutralization reaction as follows,

    $RbOH\left(aq\right)+HBr\left(aq\right)\to RbBr\left(aq\right)+H_{2}O\left(l\right)$

The acid moles and base moles consumed are determined by considering molarity and volume of the solution.

    $\begin{array}{c}Acid\text{ moles consumed = 100mL}\times \frac{1L}{1000\text{ mL}}\times \frac{0.4\text{ M HBr}}{1L}\\ \\ =\text{ 0}\text{.04 moles}\\ \\ Base\text{ moles consumed = 200mL}\times \frac{1L}{1000\text{ mL}}\times \frac{0.2\text{ M RbOH}}{1L}\\ \\ =\text{ 0}\text{.04 moles}\end{array}$

The reaction is carried under constant pressure conditions hence, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{ = q}}_{\text{p}}\text{ per mole}$.

    $\begin{array}{c}\frac{{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{1 mole HBr}}\text{= }\frac{{\text{q}}_{\text{reaction}}}{\text{moles HBr}}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}=\text{ 1 mole HBr}\times \frac{-2.23\text{ kJ}}{0.04\text{ M}}\\ \\ =\text{ -55}\text{.9 kJ}\end{array}$

Hence, the thermochemical expression is obtained as follows,

    $RbOH\left(aq\right)+HBr\left(aq\right)\to RbBr\left(aq\right)+H_{2}O\left(l\right){\Delta }_r{H}^{\circ }\text{ = -55}\text{.9 kJ}$