Question

Table 1 shows the temperature-time data were recorded for the reaction between 50.0 mL of 1.18 M HA (a weak acid) and 50.0 mL of 0.98 M NH₄OH, ammonium hydroxide (a weak base, also known as aqueous ammonia). The solutions were mixed after 60 s of approximately constant temperature readings of 24.20 ^oC. This is the initial temperature.

Table 1: temperature-time Data

Time (s)	Temperature (oC)	Time (s)	Temperature (oC)
0	24.25	90	31.22
15	24.22	105	31.12
30	24.20	120	31.02
45	24.20	135	30.92
60	24.20	150	30.82
75	31.02	165	30.72

After plotting the data using a spreadsheet, the final temperature, T_f, at the instant the second reactant was added was determined to be 31.42 ^oC. Use this data to determine ΔTΔT.

7.22 ^oC  

2. Given that the specific heat of the final salt solution is 3.92 J/g^oC and the density of the weak acid is 1.028 g/mL (the density of the NH₄OH is given in Table II of the experiment and the volumes are given in question 1), calculate the heat capacity of the system in J/^oC. [Note: the tabulated density is close enough to use even if the solution’s molarity is not exactly 1 M.] Hint: See the background section on Heat Capacities in the experiment for a useful equation.

s=q/m * Δ T

s = specific heat capacity (sometimes represented by the letter c, or Cs)

q = heat

m = mass

Δ T = change in temperature

_____J/^oC

3. Using the answers to questions 1 & 2, calculate q_cal for the reaction in kJ. Hint: The simple equation to use is the first one in the experiment. Recall that the heat capacity was calculated with units of J/oC/oC, so be sure to convert J to kJ. Hint:

Calculate Qcal

Measure the change in temperature in degrees Celsius that occurs during the reaction inside the calorimeter.

Multiply Ccal (energy/degree Celsius) by the change in temperature that occurred during the reaction in the calorimeter. For example, if the calorimeter required 3.5 Joules to increase one degree Celsius and the reaction increased the calorimeter’s temperature by 5 degrees Celsius, you would multiply 3.5 Joules/degree Celsius by 5 degrees Celsius.

Record the product of Ccal and temperature change as the total Qcal. In the example, Qcal equals 17.5 Joules, meaning that the calorimeter absorbed 17.5 Joules released by the reaction.

_____kJ

4. Given the volumes and concentrations in question 1, and that the reaction has a 1:1 stoichiometric ratio between the reactants, calculate the moles of limiting reactant in the reaction.

____mol  

5. Using the answers to questions 3 & 4, calculate the molar heat of reaction (ΔHΔH ) in kJ/mol. Remember q_cal and  ΔHΔH have opposite signs.

____kJ/mol  

Answer 1

Question 1.

The change in temperature (T) = (31.42 - 24.20) ^oC = 7.22 ^oC

Question 2.

The mass of the HA solution = (50*0.98/1.18) mL * 1.028 g/mL = 42.688 g

The mass of NH4OH solution = 50 mL * 1.028 g/mL = 51.4 g (since the acid and base almost have the same density)

Now, the mass of water that can be formed = 50 mL * 0.98 mmol/mL * 18 g/mol = 0.882 g

Now, the total mass of the solution = 46.288 g + 51.4 g - 0.882 g = 96.806 g

Therefore, the specific heat capacity (s) = 3.92 J/g.^oC * 96.806 g = 379.48 J/^oC

Note: If the density of NH4OH solution is not correct, then you can use the correct value from table II, which is not given here, by following the procedure shown above.