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Chapter 6: Problem 48

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Br}_{2}(g)\) \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{I}_{2}(g) ; \mathrm{I}_{2}(s)\)

Short Answer

Expert verified
The \(\Delta H_{\mathrm{f}}^{\circ}\) for \(Br_{2}(g)\) and \(I_{2}(g)\) is greater than zero. For \(Br_{2}(l)\) and \(I_{2}(s)\), \(\Delta H_{\mathrm{f}}^{\circ}\) is equal to zero.

Step by step solution

01

Assess Standard States of Elements

Check the given elements and their states. Recognize that for Bromine (Br) at \(25^{\circ} \mathrm{C}\), the standard state is liquid while for Iodine (I) at the same temperature, the standard state is solid.
02

Compare Given States with Standard States

Now compare the given states of the mentioned elements and their standard states at \(25^{\circ} \mathrm{C}\). For \(Br_{2}(g)\) and \(Br_{2}(l)\), the standard state is \(Br_{2}(l)\). Similarly, for \(I_{2}(g)\) and \(I_{2}(s)\), the standard state is \(I_{2}(s)\).
03

Determine \(\Delta H_{\mathrm{f}}^{\circ}\)

Based on the knowledge that the standard enthalpy of formation \(\Delta H_{\mathrm{f}}^{\circ}\) of an element in its standard state is zero, determine the \(\Delta H_{\mathrm{f}}^{\circ}\) value. For \(Br_{2}(l)\) and \(I_{2}(s)\), \(\Delta H_{\mathrm{f}}^{\circ}\) is zero as they are in their standard states. For \(Br_{2}(g)\) and \(I_{2}(g)\), \(\Delta H_{\mathrm{f}}^{\circ}\) is greater than zero because they are not in their standard states.

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Most popular questions from this chapter

How are the standard enthalpies of an element and a compound determined?

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}^{+}(a q)\right]=0\) (a) For the following reaction $$\begin{array}{r}\mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(aq)+\mathrm{Cl}^{-}(a q) \\\\\Delta H^{\circ}=-74.9 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. (b) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by 1 mole of a strong base (such as \(\mathrm{KOH}\) ) at \(25^{\circ} \mathrm{C}\).

When \(1.034 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) are burned in a constant-volume bomb calorimeter at \(298 \mathrm{~K}\), \(41.56 \mathrm{~kJ}\) of heat are evolved. Calculate \(\Delta U\) and \(\Delta H\) for the reaction on a molar basis.

Which of the following does not have \(\Delta H_{\mathrm{f}}^{\circ}=0\) at \(25^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { He }(g) & \text { Fe }(s) & \text { Cl }(g) & \text { S }_{8}(s) & \text { O }_{2}(g) & \text { Br }_{2}(l)\end{array}\)

From the following heats of combustion, $$\begin{array}{r}\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}(\mathrm{graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow\mathrm{CH}_{3} \mathrm{OH}(l)$$
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