Understanding Enthalpy: Doubling Equations and Changes in kJ/mol

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When an equation is multiplied by two, the change in enthalpy also doubles, as seen in the reaction where 2Ch_4 produces 2C and 4H_2, resulting in a ΔH of +149.8 kJ. This occurs because doubling the reactants and products requires double the energy input. However, the enthalpy change in its reduced form remains constant, as it reflects the energy change per mole. The confusion arises from the distinction between total enthalpy change and the per mole value. Ultimately, when scaling the reaction, the total enthalpy change will reflect the new stoichiometry.
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lets say u have:
Ch_4 \rightarrow C + 2H_2 \Delta H + 74.9 kJ
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
2Ch_4 \rightarrow 2C + 4H_2 \Delta H + 149.8 kJ
??
 
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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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