**Given Information:**

- ln P(solid) = –20/T 10
- ln P(liquid) = –15/T 2

**Calculating Enthalpy of Fusion:**

To calculate the enthalpy of fusion, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance in its solid and liquid states to the enthalpy of fusion.

The Clausius-Clapeyron equation is given by:

ln(P2/P1) = ΔH_fus/R * (1/T1 - 1/T2)

where:
- P1 and P2 are the vapor pressures of the substance in its solid and liquid states, respectively.
- ΔH_fus is the enthalpy of fusion.
- R is the ideal gas constant.
- T1 and T2 are the temperatures of the substance in its solid and liquid states, respectively.

**Calculating ΔH_fus:**

Let's use the given information to calculate ΔH_fus.

From the given equation, ln P(solid) = –20/T 10, we can rewrite it as:

ln(P2/P1) = –20/T 10

Comparing this equation with the Clausius-Clapeyron equation, we can see that:
- P2/P1 = e^(-20/T 10)
- T1 = 10

Similarly, from the second given equation, ln P(liquid) = –15/T 2, we can rewrite it as:

ln(P2/P1) = –15/T 2

Comparing this equation with the Clausius-Clapeyron equation, we can see that:
- P2/P1 = e^(-15/T 2)
- T2 = 2

**Substituting the values into the Clausius-Clapeyron equation:**

Now, let's substitute the values into the Clausius-Clapeyron equation and solve for ΔH_fus:

ln(P2/P1) = ΔH_fus/R * (1/T1 - 1/T2)

ln(e^(-20/T 10)) = ΔH_fus/R * (1/10 - 1/2)

-20/T 10 = ΔH_fus/R * (1/10 - 1/2)

Simplifying the equation gives:

-2 = ΔH_fus/R * (1/10 - 1/2)

-2 = ΔH_fus/R * (-4/10)

-2 = ΔH_fus/R * (-2/5)

ΔH_fus/R = 5/2

ΔH_fus = (5/2) * R

Therefore, the enthalpy of fusion, ΔH_fus, is equal to (5/2) times the ideal gas constant, R.