On Easter Sunday, April 3,1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\left.\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\right)\) b. According to The Denver Post for April 4,1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

Step by step solution

01

Find the enthalpy values for the reactants and products

To find ∆H° for the reaction, we need the standard enthalpy values for each substance involved. The problem only supplies ∆Hf° for NaNO3(aq), but for the other species, we know that the enthalpy of formation for elements in their standard state is zero. Thus, we have: ∆Hf° for \(HNO_3(aq)\) = -174.1 kJ/mol (from a data source) ∆Hf° for \(Na_2CO_3(s)\) = -1131.2 kJ/mol (from a data source) ∆Hf° for \(NaNO_3(aq)\) = -467 kJ/mol (given) ∆Hf° for \(H_2O(l)\) = -285.8 kJ/mol (from a data source) ∆Hf° for \(CO_2(g)\) = -393.5 kJ/mol (from a data source) #Step 2: Calculate the standard enthalpy change#

02

Calculate ΔH° for the reaction using Hess's law

Using Hess's law, we can calculate the reaction enthalpy change as: ∆H° = [Σ (production moles × production enthalpy)] - [Σ (reactant moles × reactant enthalpy)] ∆H° = [(2)(-467) + (1)(-285.8) + (1)(-393.5)] - [(2)(-174.1)+(1)(-1131.2)] ∆H° = -327.4 kJ/mol #Step 3: Calculate the required mass of sodium carbonate#

03

Calculate the moles of nitric acid

Given that there was 2.0 x 10^4 gallons of 70% nitric acid solution spilled. Convert the spill volume and mass into moles of nitric acid: Volume of spill = 2.0 x 10^4 gallons × 3.785 liters/gallon × 1000 cm³/liter = 7.57 x 10^7 cm³ Mass of spilled nitric acid = 1.42 g/cm³ × 7.57 x 10^7 cm³ × 70% = 7.51 x 10^7 g of HNO₃ Now, we will find the moles of nitric acid: Moles of nitric acid = mass / molar mass = (7.51 x 10^7 g) / (63.01 g/mol) = 1.19 x 10^6 mol #Step 4: Calculate the required mass of sodium carbonate#

04

Calculate the moles of sodium carbonate required

Using the stoichiometry of the balanced reaction (1 mol Na₂CO₃ neutralizes 2 mol HNO₃), we can find the needed moles of sodium carbonate: Moles of sodium carbonate = (1/2) × 1.19 x 10^6 mol = 5.94 x 10^5 mol Now, we find the mass of sodium carbonate required: Mass of sodium carbonate = 5.94 x 10^5 mol × 105.99 g/mol = 6.30 x 10^7 g #Step 5: Calculate the heat evolved during neutralization#

05

Calculate the total heat evolved

Now that we have the moles of sodium carbonate, we can find the total heat evolved during the neutralization: Total heat evolved = moles of sodium carbonate × ∆H° Total heat evolved = 5.94 x 10^5 mol × -327.4 kJ/mol = -1.94 x 10^8 kJ #Step 6: Explain the concern about air pollution#

06

Major concern about air pollution

Since the ∆H° is a large negative value, it indicates that a significant amount of heat would be released during the neutralization. This large heat release could cause the evaporation of water from the aqueous products, creating a high local concentration of airborne particulates such as `NaNO3(aq)`. Additionally, the `CO2`(g) produced might also contribute to air pollution. The major concern would likely be the potential release of high-concentration particulates and gases in the nearby atmosphere, which could be detrimental to public health.

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