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How can diamond ever be more stable than graphite, when it has

less entropy? Explain how at high pressures the conversion of graphite to diamond

can increase the total entropy of the carbon plus its environment.

Short Answer

Expert verified

As the pressure increases entropy increases.

Step by step solution

01

Given information 

Consider a system; the equilibrium state occurs when the system's entropy maximises, but the entropy of the diamond is less than the entropy of the graphite, despite the fact that the diamond is more stable. This is because the total entropy of the system plus the environment tends to increase, and in the process of Graphite to Diamond conversion under high pressure, the graphite takes up less space, leaving more space for the surrounding material, the entropy is proportional to volume.

02

Conclusion

As a result, this space permits the entropy of the surrounding material to rise. This can also be illustrated using the pressure-entropy relationship:

P=∂S∂VU

so as the pressure increases entropy increases.

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Most popular questions from this chapter

Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 x 10-6 bar-1, while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

Sulfuric acid, H2SO4,readily dissociates intoH+andHSO4-H+andHSO4-ions

H2SO4⟶H++HSO4-

The hydrogen sulfate ion, in turn, can dissociate again:

HSO4-⟷H++SO42-

The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 10 and 10*, respectively. (For dissociation of acids it is usually more convenient to look up K than ∆G°. By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = -2, while for the second reaction pK = 1.9.)

(a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?

(b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SOg. What is the pH of this rainwater?

(c) Explain why you can neglect dissociation of water into H* and OH in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO and SO2-?

The formula for CP-CV derived in the previous problem can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation P=-(∂F/∂V)T.

Use the data at the back of this book to verify the values of ∆Hand ∆G quoted above for the lead-acid reaction 5.13.

Consider the production of ammonia from nitrogen and hydrogen,

N2 + 3H2 →2NH3
at 298 K and 1 bar. From the values of â–³Hand S tabulated at the back of this book, compute â–³Gfor this reaction and check that it is consistent with the value given in the table.

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