Tag: enthalpy

Hess’s Law and Enthalpies of Formation One of the most powerful tools in thermochemistry is Hess’s Law. It allows chemists to calculate enthalpy changes for reactions that are difficult (or even impossible) to measure directly. The central idea is that enthalpy is a state function. That means the enthalpy change, $latex \Delta H$, for a reaction depends only on the initial and final states of the system, not on the path taken. What Does Hess’s Law Say? Hess’s Law states: If a reaction is carried out in a series of steps, the overall enthalpy change is the sum of the enthalpy changes for the individual steps. For example, consider the combustion of methane: $latex \text{CH}_4(g) + 2 \text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l)$$latex \Delta H = -890 \,\text{kJ}$ This overall reaction can be thought of as two steps: $latex \text{CH}_4(g) + 2 \text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad \Delta H = -802 \,\text{kJ}$ $latex 2 \text{H}_2\text{O}(g) \longrightarrow 2 \text{H}_2\text{O}(l) \quad \Delta H = -88 \,\text{kJ}$ Adding these steps gives the same overall reaction with: $latex \Delta H = -802 \,\text{kJ} + (-88 \,\text{kJ}) = -890 \,\text{kJ}$ This example shows the power of Hess’s Law: no matter which path you take, the enthalpy change is the same. Reversing and Scaling Reactions Because enthalpy is extensive, changing the equation changes $latex \Delta H$ in predictable ways: Reversing a reaction: $latex \Delta H$ changes sign. Multiplying coefficients: $latex \Delta H$ is multiplied by the same factor. Enthalpies of Formation To apply Hess’s Law systematically, chemists use tabulated values of standard enthalpies of formation, denoted $latex \Delta H_f^\circ$. This is the enthalpy change when one mole of a compound is formed from its elements in their standard states at 298 K and 1 bar. For example: $latex 2 \, C(\text{graphite}) + 3 \, H_2(g) + \frac{1}{2} O_2(g) \longrightarrow C_2H_5OH(l) \quad \Delta H_f^\circ = -277.7 \,\text{kJ/mol}$ Important Notes: The standard state of an element is its most stable form at 298 K and 1 bar (e.g., graphite for carbon, O2 for oxygen, H2 for hydrogen). By definition, $latex \Delta H_f^\circ$ for an element in its standard state is zero. Standard Enthalpies of Formation, ΔHf°, at 298 K SubstanceFormulaΔHf° (kJ/mol)SubstanceFormulaΔHf° (kJ/mol)AcetyleneC2H2(g)226.7Hydrogen chlorideHCl(g)-92.30AmmoniaNH3(g)-46.19Hydrogen fluorideHF(g)-268.60BenzeneC6H6(l)49.0Hydrogen iodideHI(g)25.9Calcium carbonateCaCO3(s)-1207.1MethaneCH4(g)-74.80Calcium oxideCaO(s)-635.5MethanolCH3OH(l)-238.6Carbon dioxideCO2(g)-393.5PropaneC3H8(g)-103.85Carbon monoxideCO(g)-110.5Silver chlorideAgCl(s)-127.0DiamondC(s)1.88Sodium bicarbonateNaHCO3(s)-947.7EthaneC2H6(g)-84.68Sodium carbonateNa2CO3(s)-1130.9EthanolC2H5OH(l)-277.7Sodium chlorideNaCl(s)-410.9EthyleneC2H4(g)52.30SucroseC12H22O11(s)-2221GlucoseC6H12O6(s)-1273WaterH2O(l)-285.8Hydrogen bromideHBr(g)-36.23Water vaporH2O(g)-241.8 Using Enthalpies of Formation to Calculate Enthalpy of Reaction The enthalpy change of any reaction under standard conditions can be calculated using the formula: $latex \Delta H_{rxn}^\circ = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})$ Here, $latex n$ and $latex m$ are the stoichiometric coefficients from the balanced chemical equation. Example: Combustion of Propane Reaction: $latex C_3H_8(g) + 5 O_2(g) \longrightarrow 3 CO_2(g) + 4 H_2O(l)$ Using standard enthalpies of formation: $latex \Delta H_{rxn}^\circ = \left[3(-393.5) + 4(-285.8)\right] - \left[(-103.85) + 0\right]$$latex = -2220 \,\text{kJ}$ This means burning 1 mol of propane releases 2220 kJ of energy as heat. Why Is Hess’s Law Useful? Hess’s Law is essential in chemistry because: It lets us calculate reaction enthalpies that are difficult to measure experimentally. It allows indirect determination of heats of combustion, vaporization, or reaction using known data. It reinforces the concept that enthalpy depends only on initial and final states, not the path taken. Example: Using Hess’s Law to Calculate Unknown Enthalpy Let’s see how Hess’s Law can be applied to determine an unknown enthalpy change. Suppose we want to calculate the enthalpy change for the formation of carbon monoxide from its elements: $latex \text{C}(s) + \frac{1}{2}\text{O}_2(g) \;\longrightarrow\; \text{CO}(g) \quad \Delta H = ?$ We are given the following thermochemical equations: $latex \text{C}(s) + \text{O}_2(g) \;\longrightarrow\; \text{CO}_2(g) \quad \Delta H = -393.5 \,\text{kJ}$ $latex \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \;\longrightarrow\; \text{CO}_2(g) \quad \Delta H = -283.0 \,\text{kJ}$ Step 1: Identify the Target Equation We want: $latex \text{C}(s) + \frac{1}{2}\text{O}_2(g) \;\longrightarrow\; \text{CO}(g)$ Step 2: Manipulate the Given Equations Equation (1) already contains $latex \text{C}(s)$ as a reactant, so we keep it as it is. Equation (2), however, has CO on the reactant side. To place CO on the product side, we reverse equation (2). When a reaction is reversed, the sign of $latex \Delta H$ is also reversed: $latex \text{CO}_2(g) \;\longrightarrow\; \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H = +283.0 \,\text{kJ}$ Step 3: Add the Two Equations Now, we combine (1) and the reversed (2): $latex \text{C}(s) + \text{O}_2(g) \;\longrightarrow\; \text{CO}_2(g) \quad \Delta H = -393.5 \,\text{kJ}$$latex \text{CO}_2(g) \;\longrightarrow\; \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H = +283.0 \,\text{kJ}$ Adding them gives: $latex \text{C}(s) + \frac{1}{2}\text{O}_2(g) \;\longrightarrow\; \text{CO}(g) \quad \Delta H = -110.5 \,\text{kJ}$ Final Answer: $latex \Delta H = -110.5 \,\text{kJ/mol}$ Example: Using Equations with Hess’s Law Sometimes, more than two equations must be combined to determine an unknown enthalpy. Consider the following problem: calculate $latex \Delta H$ for the reaction $latex 2 \, C(s) + H_2(g) \;\longrightarrow\; C_2H_2(g)$ We are given the following data: $latex C_2H_2(g) + \frac{5}{2} O_2(g) \;\longrightarrow\; 2 CO_2(g) + H_2O(l) \quad \Delta H = -1299.6 \,\text{kJ}$ $latex C(s) + O_2(g) \;\longrightarrow\; CO_2(g) \quad \Delta H = -393.5 \,\text{kJ}$ $latex H_2(g) + \frac{1}{2} O_2(g) \;\longrightarrow\; H_2O(l) \quad \Delta H = -285.8 \,\text{kJ}$ Step 1: Identify the Target Equation We want: $latex 2 \, C(s) + H_2(g) \;\longrightarrow\; C_2H_2(g)$ Step 2: Manipulate the Given Equations Equation (1) must be reversed because $latex C_2H_2$ is a product in the target reaction. Reversing changes the sign: $latex 2 CO_2(g) + H_2O(l) \;\longrightarrow\; C_2H_2(g) + \frac{5}{2} O_2(g) \quad \Delta H = +1299.6 \,\text{kJ}$ Equation (2) must be multiplied by 2, since the target has 2 carbons: $latex 2 C(s) + 2 O_2(g) \;\longrightarrow\; 2 CO_2(g) \quad \Delta H = -787.0 \,\text{kJ}$ Equation (3) remains unchanged: $latex H_2(g) + \frac{1}{2} O_2(g) \;\longrightarrow\; H_2O(l) \quad \Delta H = -285.8 \,\text{kJ}$ Step 3: Add the Equations Now, summing them: $latex (2 CO_2 + H_2O) \;\longrightarrow\; C_2H_2 + \frac{5}{2} O_2 \quad \Delta H = +1299.6 \,\text{kJ}$$latex 2 C + 2 O_2 \;\longrightarrow\; 2 CO_2 \quad \Delta H = -787.0 \,\text{kJ}$$latex H_2 + \frac{1}{2} O_2 \;\longrightarrow\; H_2O \quad \Delta H = -285.8 \,\text{kJ}$ When added, $latex 2 CO_2$, $latex H_2O$, and oxygen terms cancel appropriately, leaving: $latex 2 C(s) + H_2(g) \;\longrightarrow\; C_2H_2(g) \quad \Delta H = +226.8 \,\text{kJ}$ Final Answer: $latex \Delta H = +226.8 \,\text{kJ/mol}$ Example: Finding a Standard Enthalpy of Formation from a Reaction Enthalpy Determine the standard enthalpy of formation of solid calcium carbonate, $latex \Delta H_f^\circ[\mathrm{CaCO_3(s)}]$, using the decomposition reaction: $latex \mathrm{CaCO_3(s)} \;\longrightarrow\; \mathrm{CaO(s)} + \mathrm{CO_2(g)} \qquad \Delta H_{rxn}^\circ = +178.1\ \text{kJ/mol}$ Key relation. For any reaction at standard conditions, $latex \Delta H_{rxn}^\circ=\sum n\,\Delta H_f^\circ(\text{products})-\sum m\,\Delta H_f^\circ(\text{reactants}).$ Apply it to this reaction (all stoichiometric coefficients are 1): $latex \Delta H_{rxn}^\circ = \Delta H_f^\circ[\mathrm{CaO(s)}] + \Delta H_f^\circ[\mathrm{CO_2(g)}] - \Delta H_f^\circ[\mathrm{CaCO_3(s)}].$ Data (tabulated). $latex \Delta H_f^\circ[\mathrm{CaO(s)}] = -635.5\ \text{kJ/mol}$, $latex \Delta H_f^\circ[\mathrm{CO_2(g)}] = -393.5\ \text{kJ/mol}$. Solve for $latex \Delta H_f^\circ[\mathrm{CaCO_3(s)}]$: $latex 178.1 = (-635.5) + (-393.5) - \Delta H_f^\circ[\mathrm{CaCO_3(s)}]$$latex \Rightarrow\ \Delta H_f^\circ[\mathrm{CaCO_3(s)}] = -635.5 - 393.5 - 178.1$$latex \Rightarrow\ \Delta H_f^\circ[\mathrm{CaCO_3(s)}] = -1207.1\ \text{kJ/mol}$ Sanity check. A stable ionic solid typically has a large negative formation enthalpy, consistent with the result. Key Takeaways Hess’s Law: enthalpy is additive for multi-step reactions. $latex \Delta H_f^\circ$ values are the foundation for calculating $latex \Delta H_{rxn}^\circ$. Use the formula: $latex \Delta H_{rxn}^\circ = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})$. Always pay attention to physical states (e.g., $latex H_2O(l)$ vs $latex H_2O(g)$) because their enthalpies differ. For more information, see Hess's law | Equation, Definition, & Example, Hess's Law, and Hess's law (video) | Thermodynamics. See also: Step-by-Step Guide to Electron Configurations for AP Chemistry Success Unlock Phase Diagrams: Essential Guide to States of Matter