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Chapter 32: Problem 14

What is the value of \(\Delta H^{\circ}\) for the overall process of separating one mole of \(\mathrm{CH}_{4}\) into its constituent atoms, and then reforming one mole of \(\mathrm{CH}_{4}\) ?

Short Answer

Expert verified
The value of \(\Delta H^{\circ}\) for the entire process is zero.

Step by step solution

01

Understand the system and action

The system comprises one mole of \(\mathrm{CH}_{4}\). This is first separated into individual atoms, and then those atoms are recollected to form one mole of \(\mathrm{CH}_{4}\). Essentially, the system is returning to the same state it started from.
02

Apply the concept of state function

Remember that enthalpy (\(H\)) is a state function, which means its value relies solely on the current state of the system, irrespective of the steps taken to reach this state. Since the system starts and ends with one mole of \(\mathrm{CH}_{4}\), the total change in enthalpy (\(\Delta H^{\circ}\)) over the entire process will be zero.
03

Compute the value of \(\Delta H^{\circ}\)

Given that the initial and final states of the system are the same, the \(\Delta H^{\circ}\) for the entire process shall be zero.

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Most popular questions from this chapter

For molecules, why are all of the values for enthalpies of atom combination negative?

Is the enthalpy of atom combination for \(\mathrm{CH}_{4}(\mathrm{~g})\) exothermic or endothermic?

a) Draw the Lewis structures for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). b) Explain how these Lewis structures are consistent with the relative enthalpies of atom combination for \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\).

The net attractive potential energy in Figure 2 is \(-6.93 \times 10^{-18} \mathrm{~J}\). Will energy be required to separate the nuclei or will energy be released upon separation? Explain.

Why is \(\Delta H_{\mathrm{ac}}^{\circ}\) of \(\mathrm{C}(\mathrm{g})=0 ?\) Why is \(\Delta H_{\mathrm{ac}}^{\circ}\) of \(\mathrm{H}(\mathrm{g})=0\) ?
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