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Chapter 18: Problem 20

Find the temperatures at which reactions with the following \(\Delta H\) and \(\Delta S\) values would become spontaneous: (a) \(\Delta H=-126 \mathrm{~kJ} / \mathrm{mol}, \Delta S=84 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) (b) \(\Delta H=-11.7 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-105 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

Short Answer

Expert verified
The temperature at which the reaction (a) becomes spontaneous is approximately 1500 K, and for reaction (b) it is approximate 111.43 K.

Step by step solution

01

Identify Given Variables for (a)

For the first case (a), \(\Delta H = -126~kJ/mol = -126,000~J/mol\) and \(\Delta S = 84~J/K \cdot mol\). Note, the conversion is necessary to align the units.
02

Solve for Temperature for (a)

Apply the Gibbs Free Energy equation assuming (\(\Delta G = 0\)) to solve for the temperature when the reaction becomes spontaneous, \(T = \Delta H/\Delta S = -126,000~J/mol / 84~J/K \cdot mol\).
03

Convert J/K to K

Carry out the division with the unit conversion and round to two decimal places to get the temperature in Kelvin.
04

Identify Given Variables for (b)

For the second case (b), \(\Delta H = -11.7~kJ/mol = -11,700~J/mol\) and \(\Delta S = -105~J/K \cdot mol\). Again, ensure the units are aligned.
05

Solve for Temperature for (b)

Apply the Gibbs Free Energy equation same as Step 2 to solve for temperature, \(T = \Delta H/\Delta S = -11,700~J/mol / -105~J/K \cdot mol\). Notice the negative in the denominator would give us a positive temperature.
06

Convert J/K to K

Perform the division and round to two decimal points to get the temperature in Kelvin.

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Most popular questions from this chapter

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

Consider the decomposition of calcium carbonate. $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ Calculate the pressure in atm of \(\mathrm{CO}_{2}\) in an equilibrium process (a) at \(25^{\circ} \mathrm{C}\) and (b) at \(800^{\circ} \mathrm{C}\). Assume that \(\Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S^{\circ}=160.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) for the temperature range.

For the autoionization of water at \(25^{\circ} \mathrm{C}\), $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{w}}\) is \(1.0 \times 10^{-14}\). What is \(\Delta G^{\circ}\) for the process?
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