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For some absorption reactions, such as water absorbing sulfur dioxide, how can one determine whether the reaction is endothermic or exothermic, and by how much heat is absorbed or released? Do not give answers like: Synthesis is exothermic, while decomposition is endothermic. I want to know how to tell. This post was last edited by tongle on 2009-4-1 08:52.]
1. It is determined by measuring temperature increases and decreases through experiments. 2. Determined by consulting relevant chemical and chemical engineering materials. 3. It is determined by calculating the difference between the total energy of the reactants and the total energy of the products; if the total energy of the reactants is lower than that of the products, it is an endothermic reaction. Conversely, it is an exothermic reaction.
Judging by the enthalpy change, if it is greater than zero, heat is absorbed; if it is less than zero, heat is released. The enthalpy of various substances can be easily found in physical chemistry textbooks or textbooks on chemical thermodynamics.
Generally, the range can be determined by consulting relevant chemical and chemical engineering materials, such as the book \"Principles of Chemical Engineering\", and then calculations can be carried out using the knowledge acquired.
Simply look up the enthalpy values of the reactants and products, add up the enthalpy values of the products (making sure to multiply by the reaction coefficients), and subtract this sum from the total enthalpy of the reactants. If the resulting enthalpy difference is negative, heat is released; if it is positive, the reaction is endothermic
I think the simplest way is to provide the heat required for its reaction at the initial stage of the reaction. No further information will be provided; generally, if no additional energy is needed, it is a exothermic reaction. But such judgments generally pertain to whether it has been tested for industrial production in the laboratory; I’m not entirely sure about that. But it is a relatively convenient way to make a judgment; it’s much simpler than looking up information. This post was last edited by saturn.*ao on 2009-4-1 14:01]
It seems that the simpler the question, the more complicated it is to answer
I remember that during the sulfuric acid dilution test, the beaker became a bit hot due to the heat released during dissolution, and this could generally be felt by touching it. :Lol, something else – it’s better to check the manual, hehe
At normal temperature and pressure, it can still be determined through the enthalpy change. If it is less than zero, heat is released; otherwise, heat is absorbed.
There are many ways to do it: reaction heat, thermodynamic parameters; more directly, temperature; or it’s also possible to use general rules
It can be determined through the enthalpy change, or by examining the reaction mechanism to see if chain scission is exothermic
The most fundamental approach is to look at the enthalpy value; as for sensory aspects, I won’t go into that: lol
Theoretically, it is determined through enthalpy values; in practice, it is mainly assessed by the temperature changes during the reaction process!
What is the “reaction coefficient” here? Is it the molar amount in front of the product or reactant?
It’s not right to consider the absorption of water by sulfur dioxide or the dilution of acid as reaction heat, right? It’s better to look up data on enthalpies of formation or combustion to calculate the enthalpy of reaction.
If the heat release in the flask is not significant, it won’t be noticeable; it’s better to conduct the reaction in a reactor with insulation, as that makes the effect more apparent. If it’s unclear whether heat is being released, one can simply add an additional pipeline for the utility supply – and it’s not difficult to do
If the heat release in the flask is not significant, it won’t be noticeable; it’s better to conduct the reaction in a reactor with insulation, as that makes the effect more apparent. If it’s unclear whether heat is being released, one can simply add an additional pipeline for the utility supply – and it’s not difficult to do