Calculation of reaction heat

I won’t post a bounty notice. Could some moderator help let me know?

I won’t post a bounty notice. Could some moderator help let me know?

1. It is determined through experiments; calculations are carried out using the specific heat capacity formula: Q=cm△t, and then the heat of reaction is determined from Q based on the chemical reaction equation. 2. The heat of reaction is proportional to the amount of substance of each reactant. 3. Calculating reaction heat using bond energy: Generally, the energy absorbed to break 1 mole of a particular chemical bond is considered to be the bond energy of that bond. Bond energy is usually denoted by E, with the unit being kJ/mol. Method: ΔH = ΣE (reactants) – ΣE (products); that is, the heat of reaction equals the difference between the total bond energy of the reactants and the total bond energy of the products. Such as the reaction H2(g) + Cl2(g) → 2HCl(g) ； △H = E(H–H) + E(Cl–Cl) – 2E(H–Cl). 4. The enthalpy change of a reaction is calculated from the total energy of the reactants and products: ΔH = total energy of products – total energy of reactants. 5. Calculation based on the heat of combustion: The heat released when a substance burns is given by Q = n (fuel) × heat of combustion of that substance. 6. Calculation using Hess’s law: Hess’s law states that the heat of reaction remains the same whether it occurs in one step or in multiple steps ； In other words, the heat of reaction for a chemical reaction depends only on the initial and final states of the reaction, and not on the pathway taken. That is, if a reaction can proceed in several steps, the sum of the enthalpies of reaction for each step is equal to the enthalpy of reaction when the reaction occurs in one step. 7. Calculate using the standard molar enthalpies of formation of the reactants and products. For a reaction at a certain temperature and standard pressure, “0=ΣBVBRB” (this is a notation in which the reactants are moved to the right side of the equation by changing their signs; in this notation, the coefficients of the reactants are negative, VB is the stoichiometric coefficient for the reactant or product RB, and ΣB denotes summation over all substances). The reaction enthalpy ΔrHmθ for this reaction is equal to ΣBVBΔfHmθ(B) (as shown in the figure). θ represents the standard pressure, which is 1*10^5 Pa. In fact, this symbol is not written as “sigma”; it is simply a circle with a horizontal line through it, slightly thicker than “sigma”. “m” represents per mole of reaction), that is, the heat of reaction is equal to the algebraic sum of the products of the standard molar enthalpies of formation of all substances involved in the reaction at that state and their stoichiometric coefficients in the chemical equation. This can be proven using Hess’s law and the definition of standard molar enthalpies of formation; see Enthalpies of Formation for details. Some reference books provide the standard molar enthalpies of formation for various substances, and the heat of reaction required can be calculated by consulting them. For example, for the reaction CO(g) + H2O(g) == CO2(g) + H2(g), at 298 K and standard pressure, the standard molar enthalpies of formation for each substance are: ΔfHmθ = -110.53 kJ/mol, ΔfHmθ = -241.82 kJ/mol, ΔfHmθ = -393.51 kJ/mol, and ΔfHmθ = 0. Therefore, ΔrHmθ = Σ B_iΔfHmθ(B_i) = (-393.51*1 + 0*1 + (-110.53)*(-1) + 241.82*(-1)) kJ/mol = -41.16 kJ/mol. The heat of reaction for this reaction is -41.16 kJ/mol. 8. Calculation based on the standard molar enthalpies of combustion of the reactants and products. For many organic compounds, it is difficult to obtain them directly through the synthesis from elements; however, most organic compounds can burn, so their standard molar enthalpies of combustion are easier to determine. For the reaction in a certain state, “0=ΣBVBRB”, the enthalpy change of this reaction is also equal to △rHmθ = - ΣBVB△cHmθ(B) (as shown in the figure); this can also be proven by using Hess’s law and the definition of standard molar enthalpy of combustion. That is, the heat of reaction is equal to the negative of the algebraic sum of the products of the standard molar enthalpies of combustion of all substances involved in the reaction at that state and their stoichiometric coefficients in the chemical equation. For example, for the reaction under standard conditions: CH3CHO (l) + H2 (g) == C2H5OH (l), ΔrHmθ = -1166.37 kJ/mol, ΔcHmθ = -285.84 kJ/mol, ΔcHmθ = -1366.83 kJ/mol. Therefore, ΔrHmθ = -ΣBVBΔcHmθ(B) = -[(-1366.83)*1 + (-1166.37)*(-1) + (-285.84)*(-1)] kJ/mol = -85.38 kJ/mol. The enthalpy change for this reaction is -85.38 kJ/mol. Furthermore, the standard molar enthalpies of formation of various reactants and products can be determined based on their standard molar enthalpies of combustion and their combustion equations, which also allows for the indirect calculation of the reaction heat. Factors affecting reaction heat: Internal factors: Related to the enthalpy of formation of the reactants and products in the chemical reaction. External factors: Related to reaction temperature and pressure.

The method on the 4th floor is summarized very comprehensively

The original poster answered their own question~:o I’ll give it a try