The Student Room Group
Reply 1
Draw a Hess' diagram and notice where the arrows go

My answer:

Spoiler

Reply 2
negative ?

Mine is positive...= =...
Reply 3
Im probably wrong :smile:
What did you do to get yor answer Feth?
Reply 4
hot angel 02
calculate the enthaply change given the values of enthaply of formation
ZnCO3 -812
CO2 -393
ZnO -348
ZnCO3 (s) -> ZnO (s) + CO2(g)
would the answer be -71 kjmol-1?


enthalpy change = enthalpy formation of products - enthalpy formation of reactants

Remember to multiply by the stoichiometric coefficient as it appears in you equation, also if that value is for 2ZnCO3 for example, your final answer should be divided by 2 to give you units of energy/mol

In this case, enthalpy change = 348(1) + (-393)(1) - (-812)(1) = ?
Reply 5
Think of it like this, these are the definitions of the standard formation for each of those compounds:

1) Zn(s) + C(s) + 3O(g) --> ZnCO3(s) then ΔHf=812kJ mol1\Delta H_f = -812 \text{kJ mol}^{-1}

2) Zn(s) + O(g) --> ZnO(s) then ΔHf=348kJ mol1\Delta H_f = -348 \text{kJ mol}^{-1}

3) C(s) + 2O(g) --> CO2(g) then ΔHf=393kJ mol1\Delta H_f = -393 \text{kJ mol}^{-1}

What you're doing is essentially this:

ZnCO3(s) --> Zn(s) + C(s) + 3O(g) --> ZnO(s) + CO2(g)

Think about which way around you have to do each of those reactions (1-3) to get from the start to the finish. If you reverse the direction of a reaction, reverse the sign of the enthalpy. You should end up with a positive number (maybe think about how the bonding has changed). This reaction is entropically driven by the CO2(g) coming off at the end.

N.B. If you actually wanted to do the reactions (1-3) you'd have to go through a number of different reactions...there's no single reaction that will do any of those, they're just representative.
Reply 6
My method:

Spoiler

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