Question

When \(1 \mathrm{~mol}\) of \(\mathrm{NO}(g)\) forms from its elements, \(90.29 \mathrm{~kJ}\) of heat is absorbed. (a) Write a balanced thermochemical equation. (b) What is \(\Delta H\) when \(3.50 \mathrm{~g}\) of NO decomposes to its elements?

Short answer

The balanced thermochemical equation is \(\frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g)\).\(\text{ΔH} = -10.53\text{ kJ}\) for the decomposition of 3.50 g of NO.

Step-by-step solution

Understand the Problem

The problem discusses the formation of NO(g) from its elements and provides the amount of heat absorbed. One needs to write a thermochemical equation for the formation reaction and calculate the enthalpy change when 3.50 g of NO decomposes.

Write the Balanced Thermochemical Equation

The formation of \(\text{NO}(g)\) from its elements can be represented as: \[ \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g) \] The enthalpy change for this reaction is \(\text{ΔH} = +90.29\text{ kJ/mol}\).

Find the Molar Mass of NO

The molar mass of \(\text{NO}\) is calculated as follows: \( \text{Molar Mass of } \text{N} = 14.01\text{ g/mol} \) \[ \text{Molar Mass of } \text{O} = 16.00\text{ g/mol} \] \[ \text{Molar Mass of } \text{NO} = 14.01 + 16.00 = 30.01\text{ g/mol} \]

Convert Grams to Moles

Convert the given mass of \(\text{NO}\) into moles: \[ \text{Moles of } \text{NO} = \frac{3.50\text{ g}}{30.01\text{ g/mol}} = 0.1166\text{ mol} \]

Calculate \( \text{ΔH} \) for the Decomposition

For the decomposition reaction \[ \text{NO}(g) \rightarrow \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \] the enthalpy change \( \text{ΔH} \) is the negative of the formation reaction, so \(\text{ΔH} = -90.29\text{ kJ/mol}\). Therefore, the enthalpy change for 0.1166 moles of decomposition is: \[ \text{ΔH} = 0.1166\text{ mol} \times (-90.29\text{ kJ/mol}) = -10.53\text{ kJ} \]

Key concepts

enthalpy change

The concept of enthalpy change is crucial in understanding thermochemical equations. Enthalpy change, symbolized as \(\text{ΔH}\), represents the heat absorbed or released during a chemical reaction at constant pressure. When writing a thermochemical equation, it is important to specify whether the reaction is endothermic (absorbs heat, \(\text{ΔH} > 0\)) or exothermic (releases heat, \(\text{ΔH} < 0\)).
In the given exercise, when nitric oxide (\(\text{NO}\)) forms from nitrogen (\(\text{N}_2\)) and oxygen (\(\text{O}_2\)), the reaction absorbs 90.29 kJ of heat. This absorption of heat indicates the reaction is endothermic.
Thus, the enthalpy change for the formation of one mole of NO is \[ \text{ΔH} = +90.29 \text{ kJ/mol} \]. Understanding enthalpy changes helps in predicting how much energy will be required or released in chemical processes, which is essential for practical applications in industries such as manufacturing and pharmaceuticals.

molar mass

The molar mass of a substance is the mass of one mole of that substance. It is expressed in grams per mole (g/mol). Knowing the molar mass is crucial for converting between the mass of a substance and the amount (in moles) of that substance.
For nitric oxide (NO), we calculate the molar mass by adding the molar masses of its constituent elements:
\[ \text{Molar mass of } \text{N} = 14.01 \text{ g/mol} \] \[ \text{Molar mass of } \text{O} = 16.00 \text{ g/mol} \] \[ \text{Molar mass of } \text{NO} = 14.01 + 16.00 = 30.01 \text{ g/mol} \] This calculation shows that one mole of NO weighs 30.01 grams. When given a specific mass of NO, such as 3.50 grams, we can convert this mass to the number of moles using its molar mass:
\[ \text{Moles of No} = \frac{3.50 \text{ g}}{30.01 \text{ g/mol}} = 0.1166 \text{ mol} \] Conversions like these are foundational in chemistry for quantifying reactants and products in chemical reactions.

balanced chemical equations

A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. Balancing equations is essential for accurately representing what happens in a chemical reaction.
For the formation of nitric oxide (NO) from its elements, nitrogen gas (\( \text{N}_2 \)) and oxygen gas (\( \text{O}_2 \)), the balanced equation is:
\[ \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g) \] This equation shows that half a mole each of nitrogen and oxygen gases react to form one mole of NO. Balancing chemical equations like this one ensures that we abide by the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
During the decomposition of NO into its elements, the reverse reaction occurs:
\[ \text{NO}(g) \rightarrow \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \] In this case, the enthalpy change, \(\text{ΔH}\), will be the negative of the formation reaction, indicating an exothermic process. Balancing equations and understanding their enthalpy changes is fundamental for studying reaction energetics and stoichiometry in chemistry.