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Define the term stability from both a kinetic and a thermodynamic perspective. Give examples to show the differences in these concepts.

Short Answer

Expert verified

Kinetic stability determines reaction rate while thermodynamic stability determines whether the reaction is spontaneous or not. Common example used to describe the difference between the two is the process of diamond turning into graphite.

Step by step solution

01

Understanding the principle of thermodynamic stability and kinetic stability

When we say a reaction is kinetically stable, we are talking about how slow it is progressing with kinetic stability being inversely proportional to reaction rate.

Thermodynamic stability, on the other hand, depends on whether the reaction is spontaneous or not. If a reaction has a negative Gibbs Energy (change in free energy), that means, it is thermodynamically unstable and will occur spontaneously (and vice versa).

02

Understanding the differences in the concepts with the help of an example

The most common example is diamond turning into graphite. The reaction is thermodynamically unstable. However, it will take an extremely long time for that to happen and the reaction is therefore kinetically stable.

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Most popular questions from this chapter

What are the units for each of the following if concentrations are expressed in moles per litre and time in seconds?

a. Rate of a chemical reaction

b. Rate constant for a zero-order rate law

c. Rate constant for a first-order rate law

d. Rate constant for a second-order rate law

e. Rate constant for a third-order rate law

Consider the hypothetical reaction where the rate law is

-ฮ”[A2]ฮ”t=K[A2][B2]

The value of the rate constant at 302ยฐCis 2.45ร—10-4Lmol-1s-1, and at 508ยฐCthe rate constant is 0.891L mol-1s-1. What is the activation energy for this reaction? What is the value of the rate constant for this reaction at375ยฐC ?

Which of the following reactions would you expect to have the faster rate at room temperature? Why?

(Hint: Think of which would have the lower activation energy.)


2Ce4+(aq)+Hg22+(aq)โ†’2Ce3+(aq)+2Hg22+(aq)H3O+(aq)+OH-(aq)โ†’2H2O(l)

A certain reaction has an activation energy of 54.0kJ/mol. As the temperature is increased from 22ยฐC to a higher temperature, the rate constant increases by a factor of 7.00.Calculate the higher temperature.

Consider the hypothetical reaction

A+B+2Cโ†’2D+3E

In a study of this reaction, three experiments were run at the same temperature. The rate is defined as -d[B]/dt.

Experiment 1:

[A]0=2.0M[B]0=1.0ร—10-3M[C]0=1.0M

[B](mol/L)โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€ŠTime(s)2.7ร—10-4โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š1.0ร—1051.6ร—10-4โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š2.0ร—1051.1ร—10-4โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š3.0ร—1058.5ร—10-5โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š4.0ร—1056.9ร—10-5โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š5.0ร—1055.8ร—10-5โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š6.0ร—105

Experiment 2:

[A]0=1.0ร—10-2M[B]0=3.0M[C]0=1.0M

[A](mol/L)โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€ŠTime(s)8.9ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š1.07.1ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š3.05.5ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š5.03.8ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š8.02.9ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š10.02.0ร—10-3โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š13.0

Experiment 3:

[A]0=10.0M[B]0=5.0M[C]0=5.0ร—10-1M

[C](mol/L)โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€ŠTime(s)โ€Šโ€Šโ€Šโ€Š0.43โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š1.0ร—10-2โ€Šโ€Šโ€Šโ€Š0.36โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š2.0ร—10-2โ€Šโ€Šโ€Šโ€Š0.29โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š3.0ร—10-2โ€Šโ€Šโ€Šโ€Š0.22โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š4.0ร—10-2โ€Šโ€Šโ€Šโ€Š0.15โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š5.0ร—10-2โ€Šโ€Šโ€Šโ€Š0.08โ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Šโ€Š6.0ร—10-2

Write the rate law for this reaction, and calculate the rate constant.

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