If the ΔH for H2O(l) → H2O(s) is -6.01 kJ, then what is the ΔH for 2H2O(s) → 2H2O(l)?
A -6.01 kJ
B. 6.01 kJ
C. -12.02 kJ
D. 12.02 kJ
sure looks like D to me
twice the value, opposite direction
To determine the ΔH for the reaction 2H2O(s) → 2H2O(l), you can use the fact that ΔH is a state function. This means that the ΔH of a reaction depends only on the initial and final states, not the path taken to get there.
Given that the ΔH for H2O(l) → H2O(s) is -6.01 kJ, this means that 6.01 kJ of energy is released when liquid water freezes and becomes solid water. Since ΔH is a state function, the reverse process of vaporizing solid water to liquid water will absorb the same amount of energy, but with the opposite sign.
Therefore, the ΔH for 2H2O(s) → 2H2O(l) can be obtained by doubling the magnitude of the ΔH for H2O(l) → H2O(s). In this case, doubling -6.01 kJ gives you -12.02 kJ.
Hence, the answer is C. -12.02 kJ.