A .25 g chunk of sodium metal is cautiously dropped into a mixture of 50g of water and 50 g of ice both at 0 C
2Na(s) + 2h2o(l) -> 2naoh(aq) + H2
Delta H = -368 kj
Will the ice melt
Assuming the final mixture has a heat capacity of 4.18 J/g C calculate the final temperature
The enthalpy of fusion for ice is 6.02 kj/mol
5 answers
Is that 368,000 J for the 0.25 g Na reaction OR is it 368,000 J/mol. If 368,000 J/mol there isn't enough heat produced to melt all of the ice. If it is 368,000 J heat produced by the 0.25 g Na, there is enough heat to melt all of the ice AND raise the temperature of the resulting solution.
It is 368000J
Can You explain how work out this question?
OK but I think something is screwy. If 368,000 J for the 0.25g Na, then how much heat is required to melt all of the ice? That is q = mass ice x heat fusion (which I've changed to J/g = 334).
q = 50g ice x 334 J/g = 16,700 J so all of the ice will melt and we have heat remaining to raise the temperature of the water/melt mix.
That leaves 368,000 J-16,700 = 351,300 J to heat the water/melt mix. And that's where the problem comes in.
The final mixture mass = 100 g
Heat to raise this from zero to 100 C will be
q = 100 x 4.18 x delta T (which is 100) = 41,800
Thus the T could be raised to 100 C and use only 41,800 J more and that will leave 351,300-41,800 = 309,500 J.
That could be used to vaporize the 100 g water which will take only 2257 J/g x 100 = 225,700 J. The problem doesn't give you the 2257(40.65 kJ/mol) as it does the heat fusion which makes me think this can't be the right interpretation of the problem. (We have converted all of the ice/water to steam at 100 C and still have heat left over. You would need the specific heat steam to know how much more the temperature could rise.
If you go the other way and decide the heat is 368,000 J/mol, then you have
368,000 J x 0.25/46 = about 2000 J available. That will melt how much ice?
2000 = mass ice x 334
mass ice = about 6g; therefore you would have 50-6 = 43g ice + 56 g H2O, all at zero C as the final T.
q = 50g ice x 334 J/g = 16,700 J so all of the ice will melt and we have heat remaining to raise the temperature of the water/melt mix.
That leaves 368,000 J-16,700 = 351,300 J to heat the water/melt mix. And that's where the problem comes in.
The final mixture mass = 100 g
Heat to raise this from zero to 100 C will be
q = 100 x 4.18 x delta T (which is 100) = 41,800
Thus the T could be raised to 100 C and use only 41,800 J more and that will leave 351,300-41,800 = 309,500 J.
That could be used to vaporize the 100 g water which will take only 2257 J/g x 100 = 225,700 J. The problem doesn't give you the 2257(40.65 kJ/mol) as it does the heat fusion which makes me think this can't be the right interpretation of the problem. (We have converted all of the ice/water to steam at 100 C and still have heat left over. You would need the specific heat steam to know how much more the temperature could rise.
If you go the other way and decide the heat is 368,000 J/mol, then you have
368,000 J x 0.25/46 = about 2000 J available. That will melt how much ice?
2000 = mass ice x 334
mass ice = about 6g; therefore you would have 50-6 = 43g ice + 56 g H2O, all at zero C as the final T.
The delta h (-368kj) is for the reaction. Does the Na have any enthalpy?
0 answers