Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the labo- ratory. Set up a Hess's law cycle that will let you calculate ∆H° for the conversion of methane to ethylene: 2 CH4(g) → C2H4(g) + 2 H2(g) You can use the following information: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ∆H° = -3120.8 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H° = -890.3 kJ C2H4(g) + H2(g) → C2H6(g) ∆H° = -136.3 kJ H2O(l) ∆H°f = -285.8 kJ/mol

Guida verificata passo dopo passo

Identify the target reaction: 2 CH4(g) → C2H4(g) + 2 H2(g).

Write the given reactions and their enthalpies: (1) 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l), ΔH° = -3120.8 kJ; (2) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), ΔH° = -890.3 kJ; (3) C2H4(g) + H2(g) → C2H6(g), ΔH° = -136.3 kJ.

Reverse reaction (3) to get C2H6(g) → C2H4(g) + H2(g), changing the sign of ΔH° to +136.3 kJ.

Use reaction (2) twice to represent the combustion of 2 CH4(g), resulting in 2 CO2(g) + 4 H2O(l) and a total ΔH° of 2(-890.3 kJ).

Combine the modified reactions to form a Hess's law cycle: 2 CH4(g) → 2 CO2(g) + 4 H2O(l) (from reaction 2), 2 C2H6(g) → 4 CO2(g) + 6 H2O(l) (from reaction 1), and C2H6(g) → C2H4(g) + H2(g) (reversed reaction 3). Calculate ΔH° for the target reaction by adding the enthalpies of these steps.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to complete the reaction. This principle allows chemists to calculate the enthalpy change of a reaction by using the enthalpy changes of related reactions, making it particularly useful for reactions that are difficult to measure directly.

Video consigliato:

Percorso guidato

02:03

Hess's Law

Enthalpy of Formation (∆H°f)

The enthalpy of formation (∆H°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This value is crucial for calculating reaction enthalpies using Hess's Law, as it provides a reference point for the energy changes associated with the formation of compounds from their constituent elements.

Video consigliato:

Percorso guidato

02:34

Enthalpy of Formation

Standard Enthalpy Change (∆H°)

The standard enthalpy change (∆H°) refers to the heat change that occurs at constant pressure when reactants are converted to products under standard conditions (1 atm pressure and a specified temperature, usually 25°C). This value is essential for understanding the energy dynamics of chemical reactions and is used in Hess's Law calculations to derive the enthalpy changes for complex reactions.

Video consigliato:

Percorso guidato

02:34

Enthalpy of Formation

Pratica correlata

Domanda del libro di testo

The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?

685

views

Domanda del libro di testo

The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ

710