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
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.
The delta h (-368kj) is for the reaction. Does the Na have any enthalpy?
To determine if the ice will melt and calculate the final temperature, we need to compare the amount of heat released by the reaction with the amount of heat needed to melt the ice.
First, let's calculate the heat released by the reaction. We can use the equation q = m × ΔH, where q is the heat released, m is the mass, and ΔH is the change in enthalpy.
Given:
Mass of sodium (m) = 0.25 g
Change in enthalpy (ΔH) = -368 kJ
Converting the mass of sodium to moles:
Molar mass of sodium (Na) = 22.99 g/mol
Number of moles of Na = mass of Na / molar mass of Na
moles of Na = 0.25 g / 22.99 g/mol
Now, let's calculate the heat released by the reaction:
q = moles of Na × ΔH
q = (0.25 g / 22.99 g/mol) × -368 kJ
Next, let's calculate the heat needed to melt the ice. The enthalpy of fusion (ΔH_fusion) for ice is given as 6.02 kJ/mol.
Converting the mass of ice to moles:
Molar mass of water (H2O) = 18.02 g/mol
Number of moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 100 g / 18.02 g/mol
Now, let's calculate the heat needed to melt the ice:
q_fusion = moles of H2O × ΔH_fusion
q_fusion = (100 g / 18.02 g/mol) × 6.02 kJ
To determine if the ice will melt, we need to compare the heat released by the reaction (q) with the heat needed to melt the ice (q_fusion). If q is greater than q_fusion, the ice will melt.
If the ice melts, then the final temperature of the mixture will be the melting point of ice, which is 0°C.
To calculate the final temperature, we can use the equation q = m × c × ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of the final mixture (m) = 100 g (50 g water + 50 g ice)
Specific heat capacity (c) = 4.18 J/g °C
Initial temperature (T_i) = 0°C (since both the water and ice are at 0°C)
Final temperature (T_f) = ?
To calculate the final temperature:
q_reaction + q_fusion = m × c × (T_f - T_i)
To rearrange the equation and solve for T_f:
T_f = (q_reaction + q_fusion) / (m × c) + T_i
Now we can substitute the values and calculate the final temperature:
T_f = (q + q_fusion) / (100 g × 4.18 J/g°C) + 0°C
Plug in the previously calculated values of q and q_fusion and solve for T_f to get the final temperature.