Chapter 9, Problem 9.79PAE

A student performing a calorimetry experiment combined 100.0 ml. of 0.50 M HCI and 100.0 ml. of 0.50 M NaOH in a Styrofoam^TM cup calorimeter. Both solutions were initially at 20.0 $°$C, but when the two were mixed, the temperature rose to 23.2 $°$C

(a) Suppose the experiment is repeated in the same calorimeter but this time using 200 mL of 0.50 M HCl and 200.0 ml of 0.50 M NaOH. WIII the AT observed be greater than, less than, or equal to that in the first experiment, and why?

(b) Suppose that the experiment is repeated once again in the same calorimeter, this time using 100 mL of 1.00 M HCI and 100.0 ml. of 1.00 M NaOH. Will the $\Delta$T observed be greater than, less than, or equal to that in the first experiment, and why?

Interpretation Introduction

a.

Interpretation:

Changes in temperature in a different experiment performed in the same calorimeter but with two different solutions should be compared to the change in temperature observed in another experiment which was performed before.

Concept Introduction:

Heat released during a chemical reaction in a calorimeter is calculated by applying the formula: ${\text{ΔH=mC}}_{\text{p}}\text{ΔT}$

Here, m = mass of the solution

C_p = specific heat of the reacting solution

ΔT= change in temperature

In a Styrofoam calorimeter, the pressure is constant. Heat released at a constant pressure is known as the enthalpy of the chemical reaction.

In the first experiment, the chemical equation is:

$\text{HCl(aq)+NaOH(aq)}\to {\text{NaCl(aq)+H}}_{\text{2}}\text{O(l)}$

Total volume of the reacting solution = 100 mL + 100 mL = 200 mL

Mass of the reacting solution $\text{=}\left(\text{Density×Volume}\right)\text{=}\left(\text{1 g/mL×200 mL}\right)\text{=200 g}$

[The solution is aqueous and density of water is 1 g/mL].

Therefore,

$\begin{array}{l}{\text{ΔH=mC}}_{\text{p}}\text{ΔT}\\ \text{ΔH=}\left(\text{200 g}\right)\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)\left(\text{23}{\text{.2 }}^{\text{ο}}{\text{C-20}}^{\text{ ο}}\text{C}\right)\text{=2677}\text{.76 J}\end{array}$

At constant pressure, enthalpy of the system = 2677.76 J.

Answer to Problem 9.79PAE

Solution:

The observed $\Delta T$

value will be less than the first experiment because of the value of $\Delta T$

depends on the values of mass and specific heat of the solution. The mass of the reacting solution is more in this experiment.

Explanation of Solution

In this experiment, total volume of the reacting solution = (200 + 200) mL = 400 mL

Mass of the reacting solution $\text{=}\left(\text{Density×Volume}\right)\text{=}\left(\text{1 g/mL×400 mL}\right)\text{=400 g}$

[The solution is aqueous and density of water is 1 g/mL].

We know, $\Delta H_{sys}$ = 2677.76 J

Therefore at constant pressure,

$\begin{array}{l}{\text{ΔH=mC}}_{\text{p}}\text{ΔT}\\ \text{2677}\text{.76 J=}\left(\text{400 g}\right)\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)\left(\text{ΔT}\right)\\ \text{ΔT=}\frac{\left(\text{2677}\text{.76 J}\right)}{\left(\text{400 g}\right)\text{×}\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)}\text{=1}{\text{.6 }}^{\text{ο}}\text{C}\end{array}$

The value of $\Delta T$

in the first experiment equals to $\text{3}{\text{.2 }}^{\text{ο}}\text{C}$

Therefore, the observed $\Delta T$

value will be less than the first experiment.

Interpretation Introduction

b.

Interpretation:

Changes in temperature in a different experiment performed in the same calorimeter but with two different solutions should be compared to the change in temperature observed in another experiment which was performed before.

Concept Introduction:

Heat released during a chemical reaction in a calorimeter is calculated by applying the formula: ${\text{ΔH = mC}}_{\text{p}}\text{ΔT}$

Here, m = mass of the solution

C_p = specific heat of the reacting solution

ΔT = change in temperature

In a Styrofoam calorimeter, the pressure is constant. Heat released at a constant pressure is known as the enthalpy of the chemical reaction.

In the first experiment, the chemical equation is:

$\text{HCl(aq)+NaOH(aq)}\to {\text{NaCl(aq)+H}}_{\text{2}}\text{O(l)}$

Total volume of the reacting solution = 100 mL + 100 mL = 200 mL

Mass of the reacting solution $\text{=}\left(\text{Density×Volume}\right)\text{=}\left(\text{1 g/mL×200 mL}\right)\text{=200 g}$

[The solution is aqueous and density of water is 1 g/mL].

Therefore,

$\begin{array}{l}{\text{ΔH=mC}}_{\text{p}}\text{ΔT}\\ \text{ΔH=}\left(\text{200 g}\right)\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)\left(\text{23}{\text{.2 }}^{\text{ο}}{\text{C-20}}^{\text{ ο}}\text{C}\right)\text{=2677}\text{.76 J}\end{array}$

At constant pressure, enthalpy of the system = 2677.76 J.

Answer to Problem 9.79PAE

Solution:

The observed $\Delta T$

value will be equal to that in the first experiment because of the value of $\Delta T$

depends on the values of mass and specific heat of the solution. The mass of the reacting solution is same as that in the first experiment.

Explanation of Solution

In this experiment, total volume of the reacting solution = (100 + 100) mL = 200 mL

Mass of the reacting solution $\text{=}\left(\text{Density×Volume}\right)\text{=}\left(\text{1 g/mL×200 mL}\right)\text{=200 g}$

[The solution is aqueous and density of water is 1 g/mL].

We know, $\Delta H_{sys}$ = 2677.76 J

Therefore at constant pressure,

$\begin{array}{l}{\text{ΔH=mC}}_{\text{p}}\text{ΔT}\\ \text{2677}\text{.76 J=}\left(\text{200 g}\right)\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)\left(\text{ΔT}\right)\\ \text{ΔT=}\frac{\left(\text{2677}\text{.76 J}\right)}{\left(\text{200 g}\right)\text{×}\left(\text{4}{\text{.184 J/g/}}^{\text{ο}}\text{C}\right)}\text{=3}{\text{.2 }}^{\text{ο}}\text{C}\end{array}$

The value of ΔT in the first experiment equals to 3.2⁰C

Therefore, the observed ΔT value will be equal to that in the first experiment.