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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
It is significant to indicate the physical state of the substances in a thermochemical equation because the amount of energy that a substance absorbs or releases during a reaction can change significantly based on its physical state. This impacts the total heat change for the reaction and aids in accurately describing the energy changes that occur during the course of the reaction.

Step by step solution

01

Understanding thermochemical equations

Thermochemical equations are the chemical equations that include the enthalpy change of the reaction. They provide information about energy changes during a chemical reaction. They are written as: A + B -> C + D ΔH = X, where A and B are reactants, C and D are products, and ΔH is the enthalpy change. The equation should ideally represent the physical stages (gas, liquid, solid, or aqueous) of the reactants and products.
02

The role of the physical state

The physical state of a substance can impact how a substance reacts and thus, the enthalpy change of the reaction. The amount of energy that is absorbed or released during a chemical reaction can be drastically different depending on the states of reactant and product molecules. For example, the enthalpy of vaporization for water, which is the energy needed for water to transform from a liquid state into a gaseous state at a constant pressure, is considerably greater than the enthalpy of fusion, the energy needed to transform ice (solid water) into liquid water.
03

Conclusion

By indicating the physical state of each substance in a thermochemical equation, one can accurately describe the energy changes that occur during the course of reaction. This is because the energy required to change the state of a substance (like from solid to liquid, or liquid to gas) is included in the total heat change for the reaction.

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Most popular questions from this chapter

Consider these changes. (a) \(\operatorname{Hg}(l) \longrightarrow \operatorname{Hg}(g)\) (b) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Consider this reaction: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow & 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\ \Delta H &=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ What is the value of \(\Delta H\) if (a) the equation is multiplied throughout by \(2,\) (b) the direction of the reaction is reversed so that the products become the reactants and vice versa, (c) water vapor instead of liquid water is formed as the product?

Calculate the internal energy of a Goodyear blimp filled with helium gas at \(1.2 \times 10^{5} \mathrm{~Pa}\). The volume of the blimp is \(5.5 \times 10^{3} \mathrm{~m}^{3} .\) If all the energy were used to heat 10.0 tons of copper at \(21^{\circ} \mathrm{C},\) calculate the final temperature of the metal. (Hint: See Section 5.6 for help in calculating the internal energy of a gas. 1 ton \(\left.=9.072 \times 10^{5} \mathrm{~g} .\right)\)

Determine the standard enthalpy of formation of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) from its standard enthalpy of combustion \((-1367.4 \mathrm{~kJ} / \mathrm{mol})\)

Decomposition reactions are usually endothermic, whereas combination reactions are usually exothermic. Give a qualitative explanation for these trends.
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