   # 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 HNO 3 ( a q ) + Na 2 CO 3 ( a q ) → 2 NaNO 3 ( a q ) + H 2 O ( l ) + CO 2 ( g ) a. Calculate ∆ H° for this reaction. Approximately 2.0 ×10 4 gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0% HNO 3 by mass with a density of 1.42 glcm3. What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( Δ H f ° for NaNO 3 (aq) = −467 kJ/mol) 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 ∆ H°, what was their major concern? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 134IP
Textbook Problem
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## 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 HNO 3 ( a q ) + Na 2 CO 3 ( a q ) → 2 NaNO 3 ( a q ) + H 2 O ( l ) + CO 2 ( g ) a. Calculate ∆H° for this reaction. Approximately 2.0 ×104 gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0% HNO3 by mass with a density of 1.42 glcm3. What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( Δ H f ° for NaNO3(aq) = −467 kJ/mol)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 ∆H°, what was their major concern?

(a)

Interpretation Introduction

Interpretation: The edge length of the cube of uranium should be determined.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, substances in which all its standard states is called as a standard enthalpy of formation.
• Formula:

ΔHfnpΔH°f(products)-nrΔH°f(reactants)......(1)

• Internal energy change of a reaction is given as,

### Explanation of Solution

To calculate the enthalpy change of the neutralization reaction

Given data:

Volume of Nitric acid is 2.0 ×104gal

Nitric acid concentration is 7000

Nitric acid density is 1.42 glcm3

Enthapy of formation of NaNO3(aq) is -467 kJ/mol

Neutralization reaction is,

2HNO3(aq)+NaCO3(s)2NaNO3(aq)+H2O(l)+CO2(g)

ΔΗ°=[2(-467kJ)+(-286kJ)+(-393.5kJ)]-[2(-207kJ)+(-1131kJ)]=-69kJ

• The standard enthalpy of formation values are taking in to standard data and these are plugging in to equation (1) to give enthalpy change of the neutralization reaction.
• The enthalpy change of the neutralization reaction is -69kJ.

To calculate the mass of Nitric acid.

MassNHO3=2.0×104gallons×4qtgal×946mLqt×1.42gmL=1.1×108g

• The volume of the Nitric acid is converted by mass by converting unites for g/mL.
• The mass of Nitric acid is 1.1×108g.

To calculate the mass of 70% Nitric acid.

=MassHNO3(g)70.0gHNO3100.0gsolution=1.1×10870.0g100.0g= 7.7×107g

• The calculated mass of Nitric acid is plugging in above equation to give mass of 70% Nitric acid solution

(b)

Interpretation Introduction

Interpretation: The edge length of the cube of uranium should be determined.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, substances in which all its standard states is called as a standard enthalpy of formation.
• Formula:

ΔHfnpΔH°f(products)-nrΔH°f(reactants)......(1)

• Internal energy change of a reaction is given as,

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