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Chapter 6: Problem 131

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

Short Answer

Expert verified
Decomposition reactions are usually endothermic because they involve breaking bonds, which requires an energy input. On the other hand, combination reactions are generally exothermic, as they involve the formation of new bonds, leading to energy being released.

Step by step solution

01

Understanding Key Terms

Endothermic reactions are those that absorb heat from their surroundings. In contrast, exothermic reactions are those that release heat to their surroundings. A decomposition reaction involves breaking a compound into its constituent elements or simpler compounds. This usually requires energy input, suggesting it is endothermic. A combination reaction involves the joining of two or more elements or simpler compounds to form a more complex compound. This usually releases energy, indicating it's exothermic.
02

Describing Decomposition Reactions

In a decomposition reaction, a compound breaks down into its constituent parts. Breaking bonds requires energy input, explaining why decomposition reactions are typically endothermic. An example of a decomposition reaction is: \(2 H_2O \rightarrow 2 H_2 + O_2\). As we can see, the molecule of water is breaking down into hydrogen and oxygen molecules, which requires an input of heat energy, thus the reaction is endothermic.
03

Describing Combination Reactions

In a combination reaction, two or more elements or compounds come together to form a more complex compound. The new bonds formed result in a lowering of the system's overall energy, releasing energy. This is why combination reactions are generally exothermic. For instance, the formation of water from hydrogen and oxygen is a combination reaction: \(2 H_2 + O_2 \rightarrow 2 H_2O\). This reaction releases heat energy, so it's an exothermic process.

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Most popular questions from this chapter

On what law is the first law of thermodynamics based? Explain the sign conventions in the equation \(\Delta U=q+w\)

The standard enthalpy change for the following reaction is \(436.4 \mathrm{~kJ} / \mathrm{mol}:\) $$\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$ Calculate the standard enthalpy of formation of atomic hydrogen (H).

What is the difference between specific heat and heat capacity? What are the units for these two quantities? Which is the intensive property and which is the extensive property?

Calculate the standard enthalpy of formation for diamond, given that $$\begin{array}{l} \text { C(graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \qquad \begin{aligned} \Delta H^{\circ} &=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { diamond })+\mathrm{O}_{2}(g) \longrightarrow & \mathrm{CO}_{2}(g) \\ \Delta H^{\circ} &=-395.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} \end{array}$$

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 5.7 for help in calculating the internal energy of a gas. 1 ton \(=9.072 \times 10^{5} \mathrm{~g} .\)
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