Understanding Enthalpy: Doubling Equations and Changes in kJ/mol

[DB]

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
$$2Ch_4 \rightarrow 2C + 4H_2 \Delta H + 149.8 kJ$$
??

 

[andrewchang]

yes, it would double

 

[DB]

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...

 

[andrewchang]

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.