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Chapter 6: Problem 22

In writing thermochemical equations, why is it important to indicate the physical state (that is, gaseous, liquid, solid, or aqueous) of each substance?

Short Answer

Expert verified
The physical state of each substance in a thermochemical equation is important as it can influence the enthalpy changes associated with a reaction, and thus whether the reaction absorbs or releases heat. For instance, phase changes (from solid to liquid or gas, or vice versa) require or release heat, affecting the total heat transfer of the reaction. Hence, not specifying the physical state in a thermochemical equation could lead to incorrect interpretations of the reaction’s heat transfer.

Step by step solution

01

Understanding thermochemical equations

Thermochemical equations are a type of chemical equation that include the enthalpy change, which indicates whether the process absorbs or releases heat. They can be exothermic (releasing heat to the environment) or endothermic (absorbing heat from the environment).
02

Impact of physical states on thermochemical equations

The physical state (solid, liquid, gas, or aqueous) of a compound plays a crucial role in chemical reactions and thus in thermochemical equations, because the enthalpy changes associated with the transition between these states (known as phase changes) can significantly influence the total enthalpy change for the reaction.
03

Examples of physical state effects in thermochemical equations

For instance, heat is required to convert a solid into a liquid (melting) or a liquid into a gas (evaporation), so if a reaction product is in a higher energy state (i.e. gas) than the reactant (i.e. solid or liquid) the reaction would need to compensate by releasing more energy and thus be more exothermic. Conversely, if a product is in a lower energy state (i.e., a solid) than the reactant (i.e., a gas or liquid), the reaction would need to absorb heat and thus be endothermic. Therefore, not indicating the physical state in a thermochemical equation could lead to incorrect interpretations of the reaction’s heat transfer.

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Most popular questions from this chapter

When \(1.034 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) are burned in a constant-volume bomb calorimeter at \(298 \mathrm{~K}\), \(41.56 \mathrm{~kJ}\) of heat are evolved. Calculate \(\Delta U\) and \(\Delta H\) for the reaction on a molar basis.

State Hess's law. Explain, with one example, the usefulness of this law in thermochemistry.

Why is it dangerous to add water to a concentrated acid such as sulfuric acid in a dilution process?

In general, compounds with negative \(\Delta H_{i}^{\circ}\) values are more stable than those with positive \(\Delta H_{i}^{\circ}\) values. \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\) has a negative \(\Delta H_{\mathrm{f}}^{\circ}\) (see Table 6.4 ). Why, then, does \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\) have a tendency to decompose to \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{O}_{2}(g) ?\)

What is meant by the standard-state condition?
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