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Chapter 13: Problem 133

To prevent brain damage, a drastic medical procedure is to lower the body temperature of someone who has suffered cardiac arrest. What is the physiochemical basis for this treatment?

Short Answer

Expert verified
The physiochemical basis for treating cardiac arrest by lowering body temperature is that this intervention slows down the metabolic rate of brain cells, reducing their demand for oxygen. In conditions such as cardiac arrest, where the supply of oxygen-rich blood to the brain is compromised, therapeutic hypothermia can help prevent brain cell death and thus limit brain damage.

Step by step solution

01

Understanding the Problem

First, it's necessary to understand the conditions following cardiac arrest and how they could lead to brain damage. After cardiac arrest, the flow of oxygen-rich blood to the brain is disrupted, which can lead to brain cell death and subsequent permanent brain damage.
02

Understanding Physiochemical Processes in the Body

Biochemical reactions in the body, including in the brain, occur at a rate that is influenced by temperature. Generally, an increase in temperature speeds up these reactions, while a decrease slows them down. Therefore, when the body temperature is reduced— a process known as therapeutic hypothermia— the metabolic rate of brain cells is reduced.
03

Understanding How Lowering Body Temperature Helps

When the metabolic rate of brain cells is lowered, they require less oxygen to survive. Therefore, in the period immediately following cardiac arrest, when the supply of oxygen-rich blood to the brain may be compromised, brain cell death can be minimized through therapeutic hypothermia.
04

Linking Lowered Temperature to Therapeutic Benefit

Therapeutic hypothermia is beneficial in preserving brain tissue following cardiac arrest. By slowing the metabolic rate and thus reducing the brain's demand for oxygen, this treatment can prevent or reduce the extent of brain tissue damage caused by the lack of oxygen following cardiac arrest.

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Most popular questions from this chapter

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of "rate determining."

Many reactions involving heterogeneous catalysts are zero order; that is, rate \(=k .\) An example is the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) over tungsten (W): $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ It is found that the reaction is independent of \(\left[\mathrm{PH}_{3}\right]\) as long as phosphine's pressure is sufficiently high \((\geq 1 \mathrm{~atm}) .\) Explain.

For the reaction \(\mathrm{X}_{2}+\mathrm{Y}+\mathrm{Z} \longrightarrow \mathrm{XY}+\mathrm{XZ}\) it is found that doubling the concentration of \(\mathrm{X}_{2}\) doubles the reaction rate, tripling the concentration of Y triples the rate, and doubling the concentration of \(\mathrm{Z}\) has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of \(Z\) has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

The rate constants of some reactions double with every 10 -degree rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

Explain what is meant by the rate law of a reaction.
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