Question

A 0.250 gram chunk of sodium metal is cautiously dropped into a mixture of 50.0 grams of water and 50.0 grams of ice, both at 0 degress celsius. The reaction is:2Na(s)+2H2O(l) yields 2NaOH(aq)+H2(g) Change in H is -368 KJWill the ice melt?
Assuming the final mixture has a specific heat capacity of 4.18 J/g*degrees celsius, calculate the final temperature. The enthalpy of fusion for ice is 6.02 KJ/mol.

Answer 1

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.