Chapter 17.2, Problem 12SP

Interpretation Introduction

Interpretation: The value of heat (in kJ) released by the given reaction between two reactants when the temperature increases from ${22.5}^{\circ }\text{C}$ to ${26}^{\circ }\text{C}$ is to be calculated.

Concept introduction: The heat of a reaction, or enthalpy of a reaction, is the change in enthalpy of a chemical reaction under constant pressure. It is a thermodynamic unit helpful for determining the amount of energy per mole that is either released or created during a chemical process.

Answer to Problem 12SP

The heat released by the reaction when the temperature increased from ${22.5}^{\circ }\text{C}$ to ${26}^{\circ }\text{C}$ is $-1463\times {10}^{-3}\text{ kJ}$ .

Explanation of Solution

The $50\text{ mL}$ of water containing $0.50\text{ mol}$ HCl is mixed with $50\text{ mL}$ of water containing $0.50\text{ mol}$ NaOH at the initial temperature ${22.5}^{\circ }\text{C}$ .

The final temperature of the mixture is ${26}^{\circ }\text{C}$ .

The specific heat of the water is $4.18{\text{ J g}}^{-1}\text{C}{}^{-1}$ .

The density of water is $1.00{\text{ g mL}}^{-1}$ .

The formula to calculate the volume of a mixture is as follows:

  $V=V_{\text{HCl}}+V_{\text{NaOH}}$

Where,

• $V_{\text{HCl}}$ is the volume of water containing HCl.
• $V_{\text{NaOH}}$ is the volume of water containing NaOH.
• $V$ is the total volume of a mixture.

Substitute the values of $V_{\text{HCl}}$ and $V_{\text{NaOH}}$ in the above formula as follows:

  $\begin{array}{c}V=V_{\text{HCl}}+V_{\text{NaOH}}\\ =\text{50 mL}+\text{50 mL}\\ =\text{100 mL}\end{array}$

The total volume of the mixture is $\text{100 mL}$ .

The formula to calculate the change in temperature is as follows:

  $\Delta T=T_{\text{f}}-T_{\text{i}}$

Where,

• $T_{\text{i}}$ is the initial temperature.
• $T_{\text{f}}$ is the final temperature.
• $\Delta T$ is the change in temperature.

Substitute the values of $T_{\text{i}}$ and $T_{\text{f}}$ in the above formula as follows:

  $\begin{array}{c}\Delta T=T_{\text{f}}-T_{\text{i}}\\ ={26}^{\circ }\text{C}-{22.5}^{\circ }\text{C}\\ ={3.5}^{\circ }\text{C}\\ =\text{100 mL}\end{array}$

The change in temperature is ${3.5}^{\circ }\text{C}$ .

The formula to calculate the mass of water is as follows:

  $m=d\times V$

Where,

• $d$ is the density.
• $m$ is the mass.

Substitute the values of $d$ and $V$ in the above formula as follows:

  $\begin{array}{c}m=d\times V\\ =1.00\text{ g }\bcancel{{\text{mL}}^{-1}}\times 100\bcancel{\text{mL}}\\ =100\text{ g}\\ =\text{100 mL}\end{array}$

The mass of water is $100\text{ g}$ .

The formula to calculate the heat is as follows:

  $q=mC\Delta T$

Where,

• $q$ is the heat.
• $C$ is the specific heat.

Substitute the values of $C$ , $\Delta T$ and $m$ in the above formula as follows:

  $\begin{array}{c}q=mC\Delta T\\ =100\bcancel{\text{g}}\times 4.18\text{ J }\bcancel{{\text{g}}^{-1}\text{C}{}^{-1}}\times 3.5\bcancel{\text{C}}\\ =1463\text{ J}\end{array}$

Therefore, the heat released is $1463\text{ J}$ .

The formula to calculate the heat of a reaction is as follows:

  $\Delta H=-q$

Where,

• $\Delta H$ is the heat of a reaction.

Substitute the values of $q$ in the above formula as follows:

  $\begin{array}{c}\Delta H=-q\\ =-1463\text{ J}\end{array}$

The conversion of J into kJ is done as follows:

  $1\text{ J}={10}^{-3}\text{ kJ}$

Convert $-1463\text{ J}$ into kJ as follows:

  $\begin{array}{c}-1463\text{ J}=-1463\bcancel{\text{J}}\times \frac{{10}^{-3}\text{ kJ}}{1\bcancel{\text{J}}}\\ =-1463\times {10}^{-3}\text{ kJ}\\ =-1.463\text{ kJ}\end{array}$

Therefore, the heat of a reaction is $-1463\times {10}^{-3}\text{ kJ}$ .

Conclusion

The heat released by the reaction when the temperature increased from ${22.5}^{\circ }\text{C}$ to ${26}^{\circ }\text{C}$ is $-1463\times {10}^{-3}\text{ kJ}$ .