Understanding Enthalpy: Doubling Equations and Changes in kJ/mol

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Discussion Overview

The discussion revolves around the concept of enthalpy changes in chemical reactions, specifically addressing whether the change in enthalpy doubles when the entire reaction equation is multiplied by two. The focus includes theoretical implications and the interpretation of enthalpy in terms of kJ/mol.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes that when the equation is multiplied by two, the change in enthalpy should also double, suggesting a direct relationship between the amount of reactants/products and energy required.
  • Another participant expresses uncertainty, acknowledging that while doubling the reactants and products seems to necessitate double the energy, the enthalpy change in its reduced form should remain constant.
  • A different participant argues that the enthalpy change in the modified equation is indeed double because it reflects the total amount of energy for the doubled quantities, indicating a misunderstanding of the reduced form concept.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the relationship between the enthalpy change and the modification of the reaction equation. Some assert that the enthalpy change doubles, while others question this interpretation based on the reduced form of enthalpy.

Contextual Notes

The discussion highlights potential ambiguities in the definitions of enthalpy changes and their representations in different forms, as well as the assumptions regarding the scaling of reactions.

DB
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lets say u have:
[tex]Ch_4 \rightarrow C + 2H_2 \Delta H + 74.9 kJ[/tex]
When you multply the whole equation by 2, does the change in enthalpy double? or does it stay the same? keeping in mind that it's reduced to kJ/mol...
thanks
[tex]2Ch_4 \rightarrow 2C + 4H_2 \Delta H + 149.8 kJ[/tex]
??
 
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yes, it would double
 
sry my latex is messed up, the h should be capitalized and there should be a space in between the equation and the delta H

o so it would? how come? because I understand it as yes, when you double the amount of reactants and products you need double the energy, but even so, the enthalpy change in the equation in its reduced form, will always be the same...
 
But the enthalpy change in the second equation is double because it _isn't_ in the reduced form - it's twice the amount. otherwise, I am not sure I am understanding your question.
 

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