Chapter 8, Problem 76QAP

Consider a solution prepared by dissolving 10.00 g of NaOH in 1.00 L of water.

(a) When the solid dissolves, will the temperature of the solution increase?

(b) What is the sign of H for the process?

(c) Will dissolving 5.00 g of NaOH increase Δt?

(d) Will dissolving one mole of NaOH in 1.00 L of water increase t_f?

Interpretation Introduction

(a)

Interpretation:

An explanation must be provided for the increase or decrease in temperature when 10 g NaOH is dissolved in 1.00 L of water.

Concept introduction:

• The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

That is $\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(1)}$

where n_p and n_r are the number of moles of the products and reactants.

The values of ${\text{ΔH}}_{\text{f}}^{\text{0}}$ are measured under standard conditions of 25 °C and 1 atm pressure.

Since ΔH° is independent of pressure and varies very slightly with temperature, thus ΔH° is equal to the reaction enthalpy, ΔH.

• Chemical reactions in general involve exchange of heat between the system (reactants and products) and the surroundings. The heat (q) absorbed or evolved is given as: $\begin{array}{l}\text{q = m ×c×ΔT ------(2)}\\ \text{where:}\\ \text{m and c are the mass and specific heat of water respectively}\text{.}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}\\ {\text{where T}}_{\text{2}}{\text{ and T}}_{\text{1}}\text{ are the final and initial temperatures of the solution}\text{.}\end{array}$

Answer to Problem 76QAP

The temperature of the solution will increase.

Explanation of Solution

Given:

Mass of NaOH = 10.00 g

Volume of water = 1.00 L

Calculate the enthalpy of dissolution for 10.00 g of NaOH:

The chemical reaction corresponding to the dissolution of NaOH in water is:

$\text{NaOH(s) }\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{ Na}}^{\text{+}}{\text{(aq) + OH}}^{\text{-}}\text{(aq) }\Delta \text{H = ? kJ}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1) we have:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Na}}^{\text{+}}{\text{(aq)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(OH}}^{\text{-}}{\text{(aq))] - [1ΔH}}_{\text{f}}^{\text{0}}\text{(NaOH(s))]}\\ {\text{Substituting the ΔH}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta \text{H = [1 mole }\times (\text{-240}\text{.1 kJ/mol) + 1 mole }\times \text{(-230 kJ/mol)]-[1 mole }\times \text{(-}425.6\text{ kJ/mol)]}\\ \text{ = -240}\text{.1 kJ -230 kJ +425}\text{.6 = -44}\text{.5 kJ }\end{array}$

The enthalpy of dissolution corresponding to 1 mole of NaOH = -44.5 kJ

$\text{Moles of NaOH = }\frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}}=\frac{10.00\text{ g}}{40\text{ g/mol}}=0.25\text{ moles}$

Therefore, enthalpy of dissolution corresponding to 0.25 moles of NaOH is:

$=\frac{\text{0}\text{.25 moles NaOH }\times \text{ -44}\text{.5 kJ}}{1\text{ mole NaOH}}=-11.13\text{ kJ}$

Calculate the heat absorbed by water:

Heat absorbed by water can be deduced based on equation (2):

$\begin{array}{l}\text{q = m ×c×ΔT }\\ \text{m = mass of water }\\ \text{ = volume }\times \text{ density = 1000 ml}\times \text{ 1 g/ml = 1000 g}\\ \text{c = specific heat of water = 4}\text{.18 J/g}\text{C}\\ \text{q = 1000}\times \text{4}\text{.18}\times \Delta \text{T = 4180}\Delta \text{T}\end{array}$

Calculate the value of ΔT:

Heat lost during the dissolution of NaOH = Heat gained by water

Heat lost during the dissolution of 10.00 g NaOH = -11.13 kJ

Heat absorbed by water = 4180ΔT

$\begin{array}{l}{\text{q}}_{\text{reaction }}{\text{= - q}}_{\text{water}}\\ \text{- }11130\text{ = - 4180}\times \Delta \text{T}\\ \Delta \text{T = }\frac{11130}{4180}=2.66\text{C}\end{array}$

$\begin{array}{l}{\text{ΔT = T}}_{\text{final}}{\text{-T}}_{\text{initial}}\\ {\text{Since, ΔT is positive this implies that T}}_{\text{final}}{\text{ > T}}_{\text{initial}}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The sign of H for the dissolution of NaOH needs to be deduced.

Concept introduction:

• Chemical reactions can be broadly classified as endothermic and exothermic.

• Endothermic reactions proceed with the absorption of heat ,while an exothermic reaction involves the release of heat energy.

Answer to Problem 76QAP

The sign of H is negative.

Explanation of Solution

Dissolution is a process in which a solute dissolves in a solvent to form a solution. Here, sodium hydroxide (NaOH) is the solute and water is the solvent. From part (a), the dissolution of 10 g NaOH is accompanied by the release of 11.13 kJ of heat. Therefore, since heat is being released this is an exothermic process where the sign of enthalpy (H) is negative.

Interpretation Introduction

(c)

Interpretation:

The value of ΔT when 5 g of NaOH is dissolved needs to be determined.

Concept introduction:

• Chemical reactions in general involve exchange of heat between the system (reactants and products) and the surroundings. The heat (q) absorbed or evolved is given as: $\begin{array}{l}\text{q = m ×c×ΔT ------(2)}\\ \text{where:}\\ \text{m and c are the mass and specific heat of water respectively}\text{.}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}\\ {\text{where T}}_{\text{2}}{\text{ and T}}_{\text{1}}\text{ are the final and initial temperatures of the solution}\text{.}\end{array}$

Answer to Problem 76QAP

Dissolving 5 g of NaOH will not increase but decrease ΔT.

Explanation of Solution

The enthalpy of dissolution corresponding to 10 g of NaOH = -11.13 kJ

Therefore, enthalpy of dissolution corresponding to 5 g of NaOH is:

$=\frac{\text{5 g NaOH }\times \text{ -11}\text{.13 kJ}}{10\text{ g NaOH}}=-5.67\text{ kJ}$

Heat lost during the dissolution of NaOH = Heat gained by water

Heat lost during the dissolution of 5 g NaOH = -5.67 kJ

Heat absorbed by water = 4180 ΔT

$\begin{array}{l}{\text{q}}_{\text{reaction }}{\text{= - q}}_{\text{water}}\\ \text{- }5670\text{ = - 4180}\times \Delta \text{T}\end{array}$

$\Delta \text{T = }\frac{5670}{4180}=1.36\text{C}$

ΔT when 10 g of NaOH is dissolved = 1.36 °C

ΔT when 10 g of NaOH is dissolved = 2.66 °C (from (a) sub-part).

Therefore, dissolving 5.0 g of NaOH will decrease ΔT compared to dissolving 10.0 g of NaOH.

Interpretation Introduction

(d)

Interpretation:

The effect on the final temperature T_f when 1 mole of NaOH is dissolved in 1 L water needs to be determined.

Concept introduction:

• Chemical reactions in general involve exchange of heat between the system (reactants and products) and the surroundings. The heat (q) absorbed or evolved is given as: $\begin{array}{l}\text{q = m ×c×ΔT ------(2)}\\ \text{where:}\\ \text{m and c are the mass and specific heat of water respectively}\text{.}\\ \text{ΔT}={\text{ T}}_{\text{2}}{\text{-T}}_{\text{1}}\\ {\text{where T}}_{\text{2}}{\text{ and T}}_{\text{1}}\text{ are the final and initial temperatures of the solution}\text{.}\end{array}$

Answer to Problem 76QAP

It will increase T_f.

Explanation of Solution

The enthalpy of dissolution corresponding to 1 mole of NaOH = -44.5 kJ

Heat lost during the dissolution of NaOH = Heat gained by water

Heat absorbed by water = 4180ΔT

$\begin{array}{l}{\text{q}}_{\text{reaction }}{\text{= - q}}_{\text{water}}\\ \text{- }44500\text{ = - 4180}\times \Delta \text{T}\end{array}$

$\Delta \text{T = }\frac{44500}{4180}=10.65\text{C}$

$\begin{array}{l}{\text{ΔT = T}}_{\text{final}}{\text{-T}}_{\text{initial}}\\ {\text{Since, ΔT is positive this implies that T}}_{\text{final}}{\text{ > T}}_{\text{initial}}\end{array}$.

Also, since the value of ΔT is the highest for the dissolution of 1 mole of NaOH, this implies that the final temperature will also increase.