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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