Question

Date Section Code Data Sheet Table 11.1. Time-temperature data. A. Volume of HCl solution (mL) Molarity of HCl solution (M) Volume of NaOH solution (mL) Molarity of NaOH solution (M) Initial temperature of HCl solution (°C) Initial temperature of NaOH solution (°C) Temperatures of solution (°C) at 0.25-min. (15-sec.) intervals 0.25 Run 1 Run 2 49.5 49.8 12.119 2.119 53.2 53.6 2.087 2.087 18.9°C 18.8°C a.oc 18ợc. . 29.9 29.8 32.3 3107 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75

Answer 1

Results for Table 11.2 A

Average initial temperature of HCl and NaOH solutions

We will take the data for Run 1 to calculate this average. The average temperature will be given by:

$T_{ave}=(1/2)(T_{NaOH}+T_{HCl}) \;\;\;\;\;(1)$

Subtituting given data for T_NaOH = 19.0°C and T_HCl = 18.9°C, the average temperature is:

$T_{ave}=(1/2)(19.0+18.9)^{\circ}C=19.0^{\circ}C$

The table can be filled then

	Run 1	Run 2
Average initial temperature of HCl and NaOH solutions (°C)	19.0	18.4

Maximum temperature of HCl-NaOH solutions

Taking data from table 11.1, the maximum temperature of HCl-NaOH solutions for every run is

	Run 1	Run 2
Maximum temperature of HCl-NaOH solutions (°C)	41.2	40.0

Maximum temperature change for reaction

We can calculate the maximum temperature change for reaction by using the equation:

$\Delta T=T_{max}-T_{ave}\;\;\;\;\;(2)$

Where ΔT is the temperature change (°C), T_max is the maximum temperature of the solution from table 11.1 and T_ave is the average initial temperature calculated previously. Susbtituting data from the tables above in equation 2, we have for run 1:

$\Delta T=(41.2-19.0)^{\circ}C=22.2^{\circ}C$

And table can be filled like this

	Run 1	Run 2
Maximum temperature change for reaction (°C)	22.2	21.6

Total volume of HCl plus NaOH solutions (mL)

Total volume (V_T) is calculated as the sum of the individual volumes according to:

$V_T=V_{NaOH}+V_{HCl}\;\;\;\;\;(3)$

For run 1, total volume is:

$V_T=(49.5+53.2)mL =102.7mL$

And table can be filled like this

	Run 1	Run 2
Total volume of HCl plus NaOH solutions (mL)	102.7	103.4

Mass of final NaCl solution (g)

With the total volume of the solutions, we can assume 1 g/mL for density (d) and use it as a conversion factor and the mass of NaCl solution is given by:

$m_{NaCl}=V_T*d\;\;\;\;\;(4)$

Substituting data for run 1, the mass of NaCl solution is:

$m_{NaCl}=102.7mL*1g/mL =102.7g$

And table can be filled like this

	Run 1	Run 2
Mass of final NaCl solution (g)	102.7	103.4

Heat produced from HCl-NaOH reaction (J)

The equation used to calculate the heat (Q) produced in an aqueous reaction is:

$Q=m_{NaCl}*C*\Delta T\;\;\;\;\;(5)$

C is the heat capacity of the system, we will assume that is if formed by water with a value of C = 4.184 J/g*°C. For run 1, the heat produced is:

$Q=102.7g*\frac{4.184J}{g*^{\circ}C}*22.2^{\circ}C = 9539 J$

And table can be filled like this

	Run 1	Run 2
Heat produced from HCl-NaOH reaction (J)	9539	9345