Question

Nitrogen oxide (NO) has been found to be a key component in many biological processes. It also can react with oxygen to give the brown gas \(\mathrm{NO}_{2}\). When one mole of \(\mathrm{NO}\) reacts with oxygen, \(57.0 \mathrm{~kJ}\) of heat are evolved. (a) Write the thermochemical equation for the reaction between one mole of nitrogen oxide and oxygen. (b) Is the reaction exothermic or endothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure 8.6 is an example of such an energy diagram.) (d) What is \(\Delta H\) when \(5.00 \mathrm{~g}\) of nitrogen oxide react? (e) How many grams of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat?

Short answer

Question: Write the thermochemical equation for the reaction between nitrogen oxide (NO) and oxygen, and determine if the reaction is exothermic or endothermic. Calculate the ΔH when 5.00 g of nitrogen oxide react and find the grams of nitrogen oxide needed to liberate ten kilojoules of heat. Answer: The thermochemical equation for the reaction is: 2NO + O₂ → 2NO₂ + 57.0 kJ. The reaction is exothermic. The ΔH when 5.00 g of nitrogen oxide react is 9.50 kJ, and 5.26 g of nitrogen oxide is needed to liberate ten kilojoules of heat.

Step-by-step solution

(a) Thermochemical Equation

First, we need to write the balanced chemical equation for the reaction between nitrogen oxide (NO) and oxygen. Nitrogen dioxide (NO2) is produced: $$\mathrm{2NO} + \mathrm{O_2} \to \mathrm{2NO_2}$$ Since 57.0 kJ of heat are evolved when one mole of NO reacts, we can write the thermochemical equation as: $$\mathrm{2NO} + \mathrm{O_2} \to \mathrm{2NO_2} + 57.0\ \mathrm{kJ}$$

(b) Exothermic or Endothermic Reaction

As heat is evolved during the reaction, it means the reaction is exothermic. The given heat value (57.0 kJ) is a positive value, which further confirms the reaction to be exothermic.

(c) Energy Diagram

An energy diagram for an exothermic reaction starts with reactants having a higher energy level and moves down to the products, which have a lower energy level. The difference in energy levels corresponds to the amount of heat evolved, in this case, 57.0 kJ. Since this is an analysis question, no diagram can be provided here. Please refer to Figure 8.6 in the textbook for an example of an energy diagram.

(d) ΔH for 5.00 g of Nitrogen Oxide

In order to calculate the ΔH when 5.00 g of nitrogen oxide react, we need to first convert the mass of nitrogen oxide into moles using its molar mass: Molar mass of nitrogen oxide (NO) = 14 (N) + 16 (O) = 30 g/mol $$\text{Moles of NO} = \frac{\text{Mass of NO}}{\text{Molar mass of NO}} = \frac{5.00\ \mathrm{g}}{30\ \mathrm{g/mol}} = 0.1667\ \mathrm{mol}$$ Now, since we know that 57.0 kJ is evolved for the reaction of 1 mole of NO, we can calculate the ΔH for 0.1667 mol of NO as follows: $$\Delta H = 57.0\ \mathrm{kJ/mol} \times 0.1667\ \mathrm{mol} = 9.50\ \mathrm{kJ}$$

(e) Nitrogen Oxide Needed to Liberate Ten Kilojoules of Heat

To find the grams of nitrogen oxide required for the heat liberation of 10 kJ, first, we need to find the moles of NO, given that 1 mole of NO releases 57.0 kJ of heat: $$\text{Moles of NO} = \frac{10\ \mathrm{kJ}}{57.0\ \mathrm{kJ/mol}} = 0.1754\ \mathrm{mol}$$ Now, convert the moles of nitrogen oxide to grams using its molar mass: $$\mathrm{Mass\ of\ NO} = \mathrm{Moles\ of\ NO} \times \mathrm{Molar\ mass\ of\ NO} = 0.1754\ \mathrm{mol} \times 30\ \mathrm{g/mol} = 5.26\ \mathrm{g}$$ So, 5.26 g of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat.