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Chapter 6: Problem 40

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

Short Answer

Expert verified
The standard enthalpy of formation of an element in its standard state is always zero. Meanwhile, the standard enthalpy of formation of a compound is determined by measuring the heat exchanged during the formation of one mole of the compound from its elements, all in their standard states.

Step by step solution

01

Define Standard Enthalpy of Formation

The standard enthalpy of formation, also known as standard heat of formation, is defined as the change in enthalpy when one mole of a substance in the standard state (1 atmospheric pressure and 298.15 K) is formed from its pure elements, with all products and reactants in their standard states.
02

Standard Enthalpy of Formation for an Element

For an element in its standard state, the standard enthalpy of formation is zero because there is no formation reaction needed when the element is already in its standard state. For instance, the standard enthalpy of formation for Oxygen (O_2) in its standard state at 25°C and 1 atmospheric pressure is zero as it requires no reaction to form oxygen.
03

Standard Enthalpy of Formation for a Compound

On the other hand, the standard enthalpy of formation for a compound is measured by an experiment. It involves the formation of one mole of the compound from its elements in their standard states. The heat exchanged in the reaction, usually carried out at constant pressure and 25°C, is used to calculate the enthalpy change. This is the standard enthalpy of formation. For instance, to measure the standard enthalpy of formation for \( H_2O (l)\), the reaction \( H_2 + \frac{1}{2}O_2 \rightarrow H_2O\) would be carried out and the heat exchanged measured to get the standard enthalpy of formation.

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Most popular questions from this chapter

Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$ \begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Explain what is meant by a state function. Give two examples of quantities that are state functions and two that are not.

Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(A\) is larger than that of \(B\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

A gas expands and does \(P-V\) work on the surroundings equal to \(325 \mathrm{~J}\). At the same time, it absorbs \(127 \mathrm{~J}\) of heat from the surroundings. Calculate the change in energy of the gas.

Calculate the internal energy of a Goodyear blimp filled with helium gas at \(1.2 \times 10^{5} \mathrm{~Pa}\). The volume of the blimp is \(5.5 \times 10^{3} \mathrm{~m}^{3} .\) If all the energy were used to heat 10.0 tons of copper at \(21^{\circ} \mathrm{C},\) calculate the final temperature of the metal. (Hint: See Section 5.6 for help in calculating the internal energy of a gas. 1 ton \(\left.=9.072 \times 10^{5} \mathrm{~g} .\right)\)
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