If the ΔH for H2O(l) → H2O(s) is -6.01 kJ, then what is the ΔH for 2H2O(s) → 2H2O(l)?

A. -12.02 kJ
B. 6.01 kJ
C. -6.01 kJ
D. 12.02 kJ

I think d.

To determine the ΔH for the given reaction, we can use the fact that enthalpy changes are additive.

Given that ΔH for H2O(l) → H2O(s) is -6.01 kJ, the reverse of this process, H2O(s) → H2O(l), would have a ΔH of +6.01 kJ (since it is the opposite direction).

Since we want to find the enthalpy change for 2H2O(s) → 2H2O(l), we can multiply the ΔH value for H2O(s) → H2O(l) by 2.

2H2O(s) → 2H2O(l) will have a ΔH of 2 * (+6.01 kJ), which equals +12.02 kJ.

Therefore, the answer is D) 12.02 kJ.