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What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?

Short Answer

Expert verified
The term "lower in energy" refers to substances with less potential energy and increased stability. Liquid water is lower in energy and more stable than a mixture of hydrogen and oxygen gases due to the strong chemical bonds formed between hydrogen and oxygen atoms during the exothermic reaction.

Step by step solution

01

Understanding the term lower in energy

The term "lower in energy" refers to a system or substance with less potential energy compared to another system or substance. Lowering the energy of a system leads to increased stability. In chemistry, when a chemical reaction occurs, the molecules with lower energy are typically the more stable products and require less energy to maintain their structure.
02

Comparing energy levels

In order to compare the energy levels between a mixture of hydrogen and oxygen gases and liquid water, we need to study their chemical compositions and the energy changes associated with the formation of chemical bonds. A mixture of hydrogen and oxygen gases consists of separate H2 and O2 molecules. In this gaseous state, the molecules are moving freely and have a higher potential energy. On the other hand, liquid water consists of H2O molecules, formed when hydrogen and oxygen gases chemically react with each other. During the formation of the water molecule, two hydrogen atoms combine with one oxygen atom, releasing energy in the process.
03

Energy changes during the reaction

The chemical reaction between hydrogen and oxygen gases can be represented as: \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2O(l)}\] This reaction is exothermic, meaning it releases energy in the form of heat and light. The energy released during the formation of water molecules is mainly due to the bond formation between hydrogen and oxygen atoms, creating strong and stable chemical bonds. The energy released during this reaction lowers the potential energy of the system as a whole.
04

Determining the more stable substance

Since the formation of water molecules from hydrogen and oxygen gases is an exothermic process, the potential energy of the system decreases as a result. Liquid water has a lower potential energy than the mixture of hydrogen and oxygen gases, and thus it is considered to be "lower in energy". A substance with lower potential energy is typically more stable, as it requires less energy to maintain its structure. Therefore, liquid water is more stable than the mixture of hydrogen and oxygen gases. In conclusion, the term "lower in energy" refers to substances with less potential energy and increased stability. Liquid water has a lower potential energy and is more stable than a mixture of hydrogen and oxygen gases due to the strong chemical bonds formed between hydrogen and oxygen atoms during the exothermic reaction.

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Most popular questions from this chapter

Consider 5.5 \(\mathrm{L}\) of a gas at a pressure of 3.0 \(\mathrm{atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 \(\mathrm{L}\) . a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 \(\mathrm{L}\) . The second step results in a final volume of 10.5 \(\mathrm{L}\) . Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is 8.0 \(\mathrm{L}\) and the second step leads to a volume of 10.5 \(\mathrm{L}\) . Does the work differ from that in part b? Explain.

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)

In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

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