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Chapter 6: Problem 21

Define these terms: enthalpy, enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

Short Answer

Expert verified
Enthalpy is a property of a system, calculated as the sum of its internal energy and the product of its pressure and volume. Enthalpy of reaction is the change in enthalpy during a chemical reaction, calculated as the difference of enthalpy of the products and reactants. The heat of a reaction equals enthalpy change when the reaction occurs under constant pressure and no non-pressure-volume work is done on or by the system.

Step by step solution

01

Define Enthalpy

Enthalpy (\(H\)) is a thermodynamic property of a system. It is the sum of the internal energy (\(U\)) of the system and the product of its pressure (\(P\)) and volume (\(V\)). This can be represented by the formula: \(H = U + PV\).
02

Define Enthalpy of Reaction

Enthalpy of reaction, or heat of reaction, is the change in enthalpy of a chemical reaction that occurs at a constant pressure. It is represented as \(螖H\) and calculated as the difference in enthalpy of the products and reactants: \(螖H = H_{products} - H_{reactants}\).
03

Conditions for Heat of Reaction Equal to Enthalpy Change

The heat of a reaction (\(q\)) is equal to the enthalpy change of the reaction (\(螖H\)) under conditions of constant pressure and no non-pressure-volume work being done on or by the system. When these conditions are met, we can write: \(q_p = 螖H\).

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Most popular questions from this chapter

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$

You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$

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

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\) \(\Delta H_{\mathrm{f}}^{\circ} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}\) (rhombic) \(\longrightarrow \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (b) \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)

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 $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{NO}_{2}(g) \\ \Delta H &=-114.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$
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