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Chapter 16: Problem 60

The value of \(\Delta G^{\circ}\) for the reaction $$2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l)$$ is \(-5490 .\) kJ. Use this value and data from Appendix 4 to calculate the standard free energy of formation for \(\mathrm{C}_{4} \mathrm{H}_{10}(g).\)

Short Answer

Expert verified
The standard free energy of formation for gaseous butane (C\(_4\)H\(_{10}\)(g)) is 18.6 kJ/mol.

Step by step solution

01

Write down the given data and the equation

We are given: ∆G° for the reaction = -5490 kJ Values for the standard free energy of formation for other substances involved (from Appendix 4) To calculate the standard free energy of formation for C\(_4\)H\(_{10}\)(g), we will use the equation: ∆G° = ∆G°(products) - ∆G°(reactants)
02

Identify the standard free energy of formation values for each substance from Appendix 4

From Appendix 4: ∆Gf° of O\(_2\)(g) = 0 kJ/mol (since it is in its standard state) ∆Gf° of CO\(_2\)(g) = -394.4 kJ/mol ∆Gf° of H\(_2\)O(l) = -237.2 kJ/mol
03

Use the equation to derive the standard free energy of formation for C\(_4\)H\(_{10}\)(g)

Using the equation ∆G° = ∆G°(products) - ∆G°(reactants), substitute the given values and rearrange for the unknown value (∆Gf° of C\(_4\)H\(_{10}\)): -5490 kJ = {8 x (-394.4 kJ/mol) + 10 x (-237.2 kJ/mol)} - 2 x ∆Gf°(C\(_4\)H\(_{10}\)) - 13 x 0 Simplifying and solving for ∆Gf°(C\(_4\)H\(_{10}\)): -5490 kJ = {-3155.2 kJ/mol - 2372 kJ/mol} - 2 x ∆Gf°(C\(_4\)H\(_{10}\)) -5490 kJ = -5527.2 kJ/mol + 2 x ∆Gf°(C\(_4\)H\(_{10}\))
04

Calculate the standard free energy of formation for C\(_4\)H\(_{10}\)(g)

Now, isolate ∆Gf°(C\(_4\)H\(_{10}\)) to find its value: 2 x ∆Gf°(C\(_4\)H\(_{10}\)) = 37.2 kJ/mol ∆Gf°(C\(_4\)H\(_{10}\)) = \(\frac{37.2}{2}\) kJ/mol = 18.6 kJ/mol Thus, the standard free energy of formation for gaseous butane (C\(_4\)H\(_{10}\)(g)) is 18.6 kJ/mol.

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Most popular questions from this chapter

As \(\mathrm{O}_{2}(l)\) is cooled at 1 atm, it freezes at \(54.5 \mathrm{K}\) to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a \(y\) -intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79.

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S ?\) \(\Delta S_{\text {univ }}\) ? Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined
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