✔ 最佳答案
Zn is in excess so Ag+ is limiting.
Using the no. of moles of Ag+ for calculation instead of Zn :
Enthalpy change of the reaction
= - 903 / (0.005) (negative sign should be used as displacement reaction is exothermic)
= -180600 J mol^-1
= -180.6 kJ mol^-1
2011-04-19 19:22:53 補充：
Typo:
Using the no. of moles of Ag+ for calculation, and then using mole ratio to find out the no. of moles of Zn2+ formed:
So enthalpy change
= = - 903 / (0.005/2) (negative sign should be used as displacement reaction is exothermic)
= - 361200 J mol^-1
= - 361.2 kJ mol^-1
2011-04-19 19:22:54 補充：
Typo:
Using the no. of moles of Ag+ for calculation, and then using mole ratio to find out the no. of moles of Zn2+ formed:
So enthalpy change
= = - 903 / (0.005/2) (negative sign should be used as displacement reaction is exothermic)
= - 361200 J mol^-1
= - 361.2 kJ mol^-1
2011-04-20 13:21:27 補充：
My opinion:
I think the focus is the sign of the reaction
since displacement reaction is exothermic, negative sign should be used.
And using Zn or Ag is not the focus here.
2011-04-26 21:42:55 補充：
Yeah, specifying per mole of substance is important to determine the enthalpy change.
2011-04-26 21:42:58 補充：
In your question,
Zn(s) + 2Ag+(aq) -> Zn2+(aq) + 2Ag(s) Enthalpy change = - 361.2 kJ mol^-1
1/2 Zn(s) + Ag+(aq) -> 1/2 Zn2+(aq) + Ag(s) Enthalpy change = -180.6 kJ mol^-1
You have to specify the calculation as per mole of zinc reacted. Otherwise, the question will be quite vague.
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