Question

Data: Table 1. Reaction 1 (10.0mL of 1.5M HCl with 0.0503g Mg) Initial Temperature (HC) (C)Final Temperature (HCl + Mg) (C) 21.0 36.0 Table 2. Reaction 2 (25.0 mL of 1.5M NaOH with 20.0mL OF 1.5M HC) Initial Temperature (NaOH) (oC) 21.0 Final Temperature (NaOH + HCl) (°C) 27.0 Table 3. Reaction 3 (20.0mL of 1.5M HCI with 2.316g NaHCOs) Initial Temperature (HCD (C)Final Temperature (HCI+ NaHCO) (C) 21.0 15.5 Sample Calculations Reaction 1

Answer 1

Answer 4. Energy absorbed by or released from the surroundings.

According to the energy conservation law, the energy produced in the chemical reaction must be absorbed by the system. The net change in energy must be zero, the we apply:

We have a reaction carried out in aqueous solution, the surroundings are then water. The heat absorbed by or released from a mass water can be calculated using:

$q_{(water)}= m*C*\Delta T\: \: (1)$

Where C = 4.184 J/g*ºC is the heat capacity of water

For reaction 1, the mass of water (m) can be assumed from the volume of the HCl solution. It is to say:

Using density to convert into mass units

$m = 10.0mL*\frac{1g}{mL}= 10.0g$

We can calculate the ΔT by using:

$\Delta T = (T_f-T_i)= (36.0-21.0)^{\circ}C=15^{\circ}C$

Subtituting these values in equation (1):

$q_{(water)}=q_{(surroundings)}= 10.0g*4.184\frac{J}{g*^{\circ}C}*15^{\circ}C= 627.6J$

For reaction 2, the volume of water can be assumed as the sume of volumes of the aqueos solutions mixed. It is to say:

Using density to convert into mass units

$m = 45.0mL*\frac{1g}{mL}= 45.0g$

We can calculate the ΔT by using:

$\Delta T = (T_f-T_i)= (27.0-21.0)^{\circ}C=6^{\circ}C$

Subtituting these values in equation (1):

$q_{(water)}=q_{(surroundings)}= 45.0g*4.184\frac{J}{g*^{\circ}C}*6^{\circ}C=1,129.7J$

For reaction 3, the mass of water (m) can be assumed from the volume of the HCl solution. It is to say:

Using density to convert into mass units

$m = 20.0mL*\frac{1g}{mL}= 20.0g$

We can calculate the ΔT by using:

$\Delta T = (T_f-T_i)= (15.5-21.0)^{\circ}C=-5.5^{\circ}C$

Subtituting these values in equation (1):

$q_{(water)}=q_{(surroundings)}= 20.0g*4.184\frac{J}{g*^{\circ}C}*(-5.5^{\circ}C)= 460.2J$