A 0.770g piece of lithium metal is dropped into a mixture of 50.0g water and 50.0g ice, both at 0 degrees C.

Question

The reaction is 2Li + 2H2O = 2 LiOH + H2. Hrxn = -446 Kj/mol. Assuming no heat is lost to the surroundings, and that the final mixture has a specific heat capacity of 4.18J/g x degrees C. Calculate the final temperature in degrees C. The enthalpy fusion for ice is 6.02 Kj/mol

Answer 1

2Li + 2H₂O ==> 2 LiOH + H₂ ... Hrxn = -446 kJ/mol

First, let us find the heat generated by this reaction when 0.770 g of Li are used:

0.770 g Li x 1 mol Li / 6.94 g = 0.111 mols

0.111 mols Li x -446 kJ/mol = 49.5 kJ

Next, we can use this heat to find the final temperature of the mixture as follows:

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

The heat (q) is 49.5 kJ

Some of this heat will be used to melt the 50.0 g of ice and turn it into liquid H2O @ 0ºC. This will be a phase change with no change in temperature.

q = m∆Hfusion = 50.0 g x 1 mol / 18 g x 6.02 kJ/mol = 16.7 kJ of heat used to melt the ice

Now, we have 100 g of liquid water and we have 49.5 kJ - 16.7 kJ = 32.8 kJ of heat left over to heat this water to its final temperature.

q = mC∆T

32.8 kJ = (100 g)(0.00418 kJ/g)(∆T)

∆T = 78.5º = final temperature

(be sure to check the math)