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Chapter 6: Problem 56

The standard enthalpy change for the following reaction is \(436.4 \mathrm{~kJ} / \mathrm{mol}:\) $$\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$ Calculate the standard enthalpy of formation of atomic hydrogen (H).

Short Answer

Expert verified
The standard enthalpy of formation for atomic hydrogen (H) is 218.2 kJ/mol.

Step by step solution

01

Understand the given reaction

In the given reaction, one mole of hydrogen gas (\(H_2(g)\)) is breaking down to form hydrogen atoms. The reaction can be written as: \(H_2(g) \rightarrow 2H(g)\). The enthalpy change for this reaction is given as 436.4 kJ/mol, which is an endothermic reaction, since energy is required to break the bonds.
02

Standard enthalpy of formation calculation

The standard enthalpy of formation (∆Hf°) of any element in its standard state is zero. Since \(H_2(g)\) is in its standard state, its ∆Hf° is zero. The enthalpy change of the reaction (∆H°) is the difference between the enthalpy of formation of the products and reactants. Therefore, the standard enthalpy of formation for atomic hydrogen (H) can be calculated by the equation: ∆Hf°(H) = ∆H° - ∆Hf°(H2). After substituting the given values in this equation: ∆Hf°(H) = 436.4 kJ/mol - 0 = 436.4 kJ/mol. Thus, the standard enthalpy of formation for atomic hydrogen (H) is 218.2 kJ/mol (since the reaction produces 2 moles of atomic hydrogen)

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Most popular questions from this chapter

A 4.117 -g impure sample of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) was burned in a constant-volume calorimeter having a heat capacity of \(19.65 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). If the rise in temperature is \(3.134^{\circ} \mathrm{C},\) calculate the percent by mass of the glucose in the sample. Assume that the impurities are unaffected by the combustion process. See Appendix 2 for thermodynamic data.

These are various forms of energy: chemical, heat, light, mechanical, and electrical. Suggest ways of interconverting these forms of energy.

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 M\) AgNO \(_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure \(6.9 .\) As a result of the reaction $$\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)$$ the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the above reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$\begin{array}{r}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\\\Delta H=-114.6 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

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 \(25^{\circ} \mathrm{C}\).
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