3. A student determined the Hneutzn of nitric acid (HNO3) mixed with NaOH soluti

Question

3. A student determined the Hneutzn of nitric acid (HNO3) mixed with NaOH solution, using the procedure described in this experiment. Fifty milliliters of 1.00M HNO3 was added to 50.5 mL of 1.00M NaOH solution, and the following time-temperature data were recorded temp., C 31.2 31.1 31.0 30.9 30.8 30.7 time, min temp time, mirn 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 0.0 1.0 2.0 24.7 24.8 24.9 mix 4.0 5.0 6.0 7.0 8.0 30.2 31.2 31.5 31.5 31.4 31.2 31.2 30.6 30.5 30.4 30.3 9.0 10.0 (1) Make a plot of these time-temperature data.

Explanation / Answer

1) Plot the data.

2) The final temperature of the mixture (right after mixing) is calculated by extrapolation of S3 above. Put x = 0 and obtain

y = -0.0772*(0.0) + 31.934

====> y = 31.934

The final temperature of the mixture is 31.934°C and the change in temperature, T = (final temperature) – (initial temperature) = (31.934°C) – (25°C) = 6.934°C (ans).

3) Volume of the reaction mixture = (volume of HNO3) + (volume of NaOH) = (50.0 mL) + (50.5 mL) = 100.5 mL (ans).

4) Mass of the reaction mixture = (volume of the reaction mixture)*(density of the reaction mixture) = (100.5 mL)*(1.03 g/mL) = 103.515 g (ans).

5) The heat transferred for the reaction = (mass of the reaction mixture)*(heat capacity of the mixture)*T = (103.515 g)*(3.89 J/g.°C)*(6.934°C) = 2792.137 J (ans).

6) Moles of HNO3 = (volume of HNO3 in L)*(molar concentration of HNO3) = (50.0 mL)*(1 L/1000 mL)*(1.00 M)*(1 mol/L/1 M) = 0.05 mole.

Moles of NaOH = (volume of NaOH in L)*(molar concentration of NaOH) = (50.5 mL)*(1 L/1000 mL)*(1.00 M) = 0.0505 mole.

7) In the given reaction, we have an excess of NaOH. The balanced chemical equation is

HNO3 (aq) + NaOH (aq) --------> NaNO3 (aq) + H2O (l)

As per the stoichiometric equation,

1 mole HNO3 = 1 mole NaOH = 1 mole H2O

Since we have NaOH in excess, HNO3 is the limiting reactant; the number of moles of H2O formed = moles of HNO3 = 0.05 mole.

Hneutzn = -(heat transferred for the reaction)/(moles of H2O formed) = -(2792.137 J)/(0.05 mole) = -55842.74 J/mol = -(55842.74 J/mol)*(1 kJ/1000 J) = -55.84274 kJ/mol -55.8 kJ/mol (ans).

8) Percent error = [-58.5 kJ/mol – -55.8 kJ/mol]/[-58.5 kJ/mol]*100 = 4.615% 4.6% (ans).

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.