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Chapter 22: Problem 169

Stretch a rubber band while holding it gently to your lips. Then slowly let it relax while still in contact with your lips. a. What happens to the temperature of the rubber band on stretching? b. Is the stretching an exothermic or endothermic process? c. Explain the above result in terms of intermolecular forces. d. What is the sign of \(\Delta S\) and \(\Delta G\) for stretching the rubber band? e. Give the molecular explanation for the sign of \(\Delta S\) for stretching.

Short Answer

Expert verified
When you stretch a rubber band and hold it gently to your lips, you can feel that the temperature of the rubber band increases, meaning it becomes warmer on stretching. This is an exothermic process, as heat is being released to its surroundings. The increase in temperature can be explained by the intermolecular forces (like van der Waals forces) being disrupted, which releases energy in the form of heat. When stretching the rubber band, the polymer chains align and the system becomes more ordered, resulting in a negative change in entropy (\(\Delta S\)). The sign of the Gibbs free energy change (\(\Delta G\)) depends on the magnitude of \(T\Delta S\) compared to \(\Delta H\). The decrease in entropy is due to the polymer chains in the rubber band becoming more aligned when stretched, reducing the degree of randomness.

Step by step solution

01

a. Change in temperature during stretching

When you stretch the rubber band and hold it gently to your lips, you can feel that the temperature of the rubber band increases. This means that the rubber band becomes warmer on stretching.
02

b. Exothermic or Endothermic process

An exothermic process releases heat to its surroundings, while an endothermic process absorbs heat from its surroundings. Since the rubber band becomes warmer during the stretching process, this indicates that it is releasing heat to its surroundings. Therefore, the stretching of the rubber band is an exothermic process.
03

c. Intermolecular forces

The increase in temperature during the stretching process can be explained by considering the intermolecular forces in the rubber band. When the rubber band is stretched, the polymer chains are pulled apart, and the intermolecular forces (like van der Waals forces) between the chains are overcome. As the intermolecular forces are disrupted, energy is released in the form of heat, which causes the increase in temperature.
04

d. Sign of \(\Delta S\) and \(\Delta G\) for stretching

The change in entropy (\(\Delta S\)) is positive when a system becomes more disordered and negative when it becomes more ordered. Stretching a rubber band aligns the polymer chains in the rubber, making it more ordered. Thus, the sign of \(\Delta S\) for stretching the rubber band is negative. The Gibbs free energy change (\(\Delta G\)) can be calculated using the equation: \[\Delta G =\Delta H - T\Delta S\] Here, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature, and \(\Delta S\) is the change in entropy. Since the stretching process is exothermic, \(\Delta H\) is negative. As we determined earlier, \(\Delta S\) is also negative for the stretching process. Therefore, the sign of \(\Delta G\) depends on the magnitude of \(T\Delta S\) compared to \(\Delta H\). If \(T\Delta S\) is greater than or equal to \(\Delta H\), then \(\Delta G\) will be positive or zero, making the process non-spontaneous or in equilibrium. If \(T\Delta S\) is less than \(\Delta H\), then \(\Delta G\) will be negative, indicating a spontaneous process.
05

e. Molecular explanation for the sign of \(\Delta S\) for stretching

As mentioned earlier, \(\Delta S\) is negative for the stretching process because the rubber band becomes more ordered. The molecular explanation for this behavior can be given as follows: In the relaxed state, the polymer chains in the rubber band are randomly coiled and entangled, which corresponds to a high degree of randomness or disorder. When the rubber band is stretched, these polymer chains become more aligned and the degree of randomness decreases. This results in a decrease in entropy, and therefore, the sign of \(\Delta S\) for stretching is negative.

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