Question

Calculate \(\Delta H^{\circ}\) in kilojoules for the following reaction, the preparation of the unstable acid nitrous acid, \(\mathrm{HNO}_{2}\). $$ \mathrm{HCl}(g)+\mathrm{NaNO}_{2}(s) \longrightarrow \mathrm{HNO}_{2}(l)+\mathrm{NaCl}(s) $$ Use the following thermochemical equations \(2 \mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HCl}(g)+\mathrm{Na}_{2} \mathrm{O}(s)\) \(\Delta H^{\circ}=+507.31 \mathrm{~kJ}\) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)\) $$ \begin{aligned} \Delta H^{\circ} &=-427.14 \mathrm{~kJ} \\ \mathrm{NO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) & \\ \Delta H^{\circ} &=-42.68 \mathrm{~kJ} \end{aligned} $$ \(2 \mathrm{HNO}_{2}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=+34.35 \mathrm{~kJ}\)

Short answer

-663.62 kJ

Step-by-step solution

Analyze the Given Equations

We need to re-arrange the given equations so that we can add them to get the target equation, which is the formation of nitrous acid (HNO2). The target process is: HCl(g) + NaNO2(s) -> HNO2(l) + NaCl(s).

Reverse the First Equation

We will reverse the first equation because we need HCl(g) as a reactant and NaCl(s) as a product. When we reverse a thermochemical equation, we must also reverse the sign of the enthalpy change. So, the reversed first equation will be: 2 HCl(g) + Na2O(s) -> 2 NaCl(s) + H2O(l) with \( \Delta H^\circ = -507.31 \text{ kJ} \).

Use Half the Quantities in the First Equation

Since we only need one mole of HCl(g) and NaCl(s), we will use half of the reversed first equation. This means we will also take half the enthalpy change: \( \Delta H^\circ = -507.31 \text{ kJ}/2 = -253.655 \text{ kJ} \).

Use the Second Equation As-is

The second equation is already in the correct direction, so we will use it as it is, with its given \( \Delta H^\circ = -427.14 \text{ kJ} \).

Use Half the Fourth Equation

The fourth equation needs to be halved because we are looking for the formation of one mole of HNO2(l). Hence, the halved fourth equation is: \( \mathrm{HNO}_{2}(l) \longrightarrow \frac{1}{2} \mathrm{N}_{2} \mathrm{O}(g) + \frac{1}{2} \mathrm{O}_{2}(g) + \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(l) \) with \( \Delta H^\circ = +34.35 \text{ kJ}/2 = +17.175 \text{ kJ} \).

Add the Adjusted Equations

Add the enthalpy changes from Steps 3, 4, and 5: \( (-253.655) + (-427.14) + 17.175 = \Delta H^\circ \).

Calculate the Final Enthalpy Change

Sum the values of \( \Delta H^\circ \) for all reactions to determine the enthalpy change of the target reaction: \( (-253.655) + (-427.14) + 17.175 = -663.62 \text{ kJ} \).

Key concepts

Thermochemistry

In the fascinating world of thermochemistry, we study the energy and heat associated with chemical reactions and physical transformations. The key player in these processes is energy in the form of heat, and a crucial concept is the enthalpy of a system. Enthalpy is a measure of the total heat content in a chemical system at constant pressure. Understanding enthalpy is vital, as it helps scientists and engineers determine how much energy is absorbed or released during a reaction.
Enthalpy changes can be either endothermic or exothermic. In an endothermic reaction, the system absorbs heat from the surroundings, while in an exothermic reaction, it releases heat. Calculating these enthalpy changes allows chemists to predict the direction and extent of chemical processes, crucial for designing industrial chemical reactions or energy-efficient processes.

Enthalpy of Reaction

The enthalpy of reaction, or reaction enthalpy, symbolized as \( \Delta H \), is the change in enthalpy during a chemical reaction at constant pressure. It's what scientists use to quantify the heat either absorbed or evolved in a reaction. This value can be directly related to the breaking and forming of chemical bonds, with bond breakage absorbing energy (endothermic) and bond formation releasing energy (exothermic).
In a homework problem or a laboratory setting, you would often calculate this enthalpy change using thermochemical equations provided for various steps of the reaction. By carefully adjusting these steps, as seen in our exercise, and taking into account stoichiometry and the laws of thermodynamics, students can determine the overall enthalpy change for even complex multi-step chemical reactions.

Hess's Law

Hess's Law is akin to a superpower in thermochemistry, allowing you to calculate enthalpy changes for reactions that are tough to measure directly. This principle asserts that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. In other words, the enthalpy change is path-independent.
When faced with an equation that does not directly give you the enthalpy change, Hess's Law comes to the rescue. It allows you to manipulate and sum up the enthalpies of a series of reactions to arrive at the overall enthalpy change for your target reaction. The key is ensuring that the intermediate species cancel out, just like in algebra, to derive the target equation, as demonstrated in the exercise provided.

Chemical Thermodynamics

At the intersection of chemistry and physics, we find chemical thermodynamics, the study of energy transformations within chemical reactions. The main focus is understanding the transfer of energy as heat and work, and the consequences it has on the properties of substances. The laws of thermodynamics form the framework for this field, guiding principles that dictate energy conservation, the directionality of processes, and the limits of energy conversions in chemical processes.
Within the context of chemical thermodynamics, the calculation of enthalpy is a practical application that resonates deeply with energy considerations in real-world scenarios. This involves not just the heat of reaction but also the understanding of how energy conservation impacts chemical equilibrium, reaction spontaneity, and the efficiency of energy use in chemical reactions.