Total enthalpy change for freezing of 1 mol of water at 10 C to ice at -10 C is (Given: fusH = x kJ/mol Cp[H2O(l)] = y J mol-1 K-1 Cp[H2O(s)] = z J mol-1 K-1)
- -x-10y-10z
- x-10y-10z
- -10(100x+y+z)
- 10(100x+y+z)
Answer
(3) -10(100x+y+z)
Solution
Step 1: H2O(l), 10 C H2O(l), 0 C
H1 = Cp T = y(0-10) = -10y J
Step 2: H2O(l), 0 C H2O(s), 0 C (freezing is the reverse of fusion)
H2 = - fusH = -x kJ = -1000x J
Step 3: H2O(s), 0 C H2O(s), -10 C
H3 = z(-10-0) = -10z J
Htotal = H1 + H2 + H3
= -1000x - 10y - 10z
= -10(100x + y + z) J
Answer: (3)