Question #899a4

1 Answer
anor277
Apr 25, 2017

#DeltaH_"combustion"^@=-242*kJ*mol^-1#.

Explanation:

#DeltaH^@# values are ALWAYS quoted per mole of reaction as written.

And thus:

#2H_2(g) +O_2(g) rarr 2H_2O(g)# #DeltaH^@=-484*kJ*mol^-1#.

equivalently,

#H_2(g) +1/2O_2(g) rarr 2H_2O(g)# #DeltaH^@=-242*kJ*mol^-1#.

Now, by definition, #DeltaH_"combustion"^@# is the enthalpy change when ONE mole of substance undergoes combustion under standard conditions.

And thus for #"dihydrogen"#, #DeltaH_"combustion"^@=-242*kJ*mol^-1#.

And if we combust a #1*g# mass of #H_2#, i.e. a half molar quantity with respect to dihydrogen, the energy released will reflect the molar quantity, i.e. #DeltaH^@=-141*kJ#.

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