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Zinc dissolves in acid according to the balanced equation. Zn(s)+ 2H(aq)-> Zn(aq) + H2(g) a sample of zinc is placed in a ice c...Asked by Kjd
Zinc dissolves in acid according to the balanced reaction:
Zn(s) + 2H^+(aq) -> Zn^2+(aq) + H2(g)
A sample of zinc is placed in the ice calorimeter. If 0.0657g of zinc causes a decrease of 0.109mL in the ice/water volume of the calorimeter, what is the enthalpy change, per mole of zinc, for the above reaction per mole of zinc.
Extra info:
Given that the enthalpy change for the process H2O(s) -> H2O(l) is +6.01 kJ/mol or 333 J/g, it can be determined that 3.68 kJ are released per mL change in the volume of the ice/water mixture.
This is what I have so far..
N(zinc)= m(zinc)/M(zinc) = 0.001 mol Zn
q(calorimeter)= -3.68 kJ/mL x 0.109mL
= -0.401 kJ
-q(rxn) = q(calorimeter)
q(rxn)= 0.401 kJ
(delta)H = 0.401 kJ/0.001 mol
= 401 kJ/mol
The answer is right, it's just that I do not understand why we would use 3.68 kJ/mL and multiply it with 0.109 mL?
Please help, much appreciated. Thank you!
Zn(s) + 2H^+(aq) -> Zn^2+(aq) + H2(g)
A sample of zinc is placed in the ice calorimeter. If 0.0657g of zinc causes a decrease of 0.109mL in the ice/water volume of the calorimeter, what is the enthalpy change, per mole of zinc, for the above reaction per mole of zinc.
Extra info:
Given that the enthalpy change for the process H2O(s) -> H2O(l) is +6.01 kJ/mol or 333 J/g, it can be determined that 3.68 kJ are released per mL change in the volume of the ice/water mixture.
This is what I have so far..
N(zinc)= m(zinc)/M(zinc) = 0.001 mol Zn
q(calorimeter)= -3.68 kJ/mL x 0.109mL
= -0.401 kJ
-q(rxn) = q(calorimeter)
q(rxn)= 0.401 kJ
(delta)H = 0.401 kJ/0.001 mol
= 401 kJ/mol
The answer is right, it's just that I do not understand why we would use 3.68 kJ/mL and multiply it with 0.109 mL?
Please help, much appreciated. Thank you!
Answers
Answered by
anon
3.68kj of energy are released per ml so it is actually 3.68kj/ml you need to cancel out the ml by multiplying to get the heat lost by ice water for this specific amount of ice water
Answered by
Kjd
Thanks!
Answered by
Frank
Because we need to Cancel the mL and keep kj our main unit and the heat is released as a form of energy which is measured in Kj so basically u are multiplying these 2 numbers just for the sake of getting a pure q(calorimeter) in its energy form i.e Kj because ML is not an energy unit its used to measure volume and when u use volume ml is not used often it is mostly converted into Liters (L). I wish this helped!
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