Video Transcript
The standard enthalpy change of
fusion can be described by this equation: H2O solid reacts to form H2O liquid, Δ𝐻
fusion is equal to positive 6.01 kilojoules per mole. Using this information, determine
the standard enthalpy change for two H2O liquid reacts to form two H2O solid.
The standard enthalpy of fusion is
the enthalpy change when one mole of a substance transforms from a solid state to a
liquid state under standard conditions. This question gave us the standard
enthalpy of fusion for water. So this is the transformation from
ice to liquid water, which is the process we usually call melting. The equation we are interested in
in this question has the enthalpy change for the transformation from liquid to
solid, in other words, going from water to ice. We usually refer to this process as
freezing.
The enthalpy change for this
process is called the standard enthalpy of solidification. We can visualize the relationship
between the enthalpy of fusion and solidification on this energy-level diagram. The enthalpy change going from a
solid to a liquid is the standard enthalpy of fusion. The enthalpy change going the other
way, from a liquid to a solid, is the standard enthalpy of solidification.
We can see that these two enthalpy
changes on the energy diagram have the same magnitude but opposite signs. So the standard enthalpy of
solidification for water must be equal to negative 6.01 kilojoules per mole. The equation in this question has
two moles of liquid and solid water. But the standard enthalpy of
solidification is for when one mole of a substance transforms, not two.
When we want to know the enthalpy
change of a process that is multiplied by a coefficient, we simply multiply the
enthalpy change by that coefficient. If the coefficient is two, like in
this case, we’ll multiply the enthalpy change by two. So the enthalpy change for the
equation in this question is two times the standard enthalpy of solidification for
water. So if we multiply negative 6.01 by
two, we’ll get our answer. That calculation gives us negative
12.02 kilojoules per mole.
So the standard enthalpy change for
the process two H2O liquid reacts to form two H2O solid is negative 12.02 kilojoules
per mole.
