# 16.1 Spontaneity

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By the end of this section, you will be able to:
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes

In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur by force. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions and how relatively quickly or slowly that natural change proceeds. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.

## Spontaneous and nonspontaneous processes

Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth’s atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. A spontaneous process    is one that occurs naturally under certain conditions. A nonspontaneous process    , on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.

The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years ( [link] ).

As another example, consider the conversion of diamond into graphite ( [link] ).

$\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{diamond}\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)$

The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.

a student was required to prepare 500cm of 1.0M solution of glucose(C6H12O6). (C=12, H=1, O=16).determine a.molar mass of glucose b.amount of glucose in mole c.mass of glucose contained in the solution
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ln(k two/k one)= (- activation energy/R)(1/T two - 1/T one) T is tempuratue in Kelvin, K is rate constant but you can also use rate two and one instead, R is 8.314 J/mol×Kelvin
Drenea
if ur solving for k two or k one you will need to use e to cancel out ln
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olotu
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olotu
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hey I don't understand anything in chemistry so I was wondering if you could help me
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Hi am Isaac, The number of electrons within the outer shell of the element determine its valency . To calculate the valency of an element(or molecule, for that matter), there are multiple methods. ... The valency of an atom is equal to the number of electrons in the outer shell if that number is fou
YAKUBU