Enthalpy change for a van der waal gas
The equation for $\Delta H$ for a van der Waals gas should be:
$$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes).  For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$

Explanation:
In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by:  $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$
where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).

The enthalpy H of the gas is equal to U + PV:
$$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$
Combining the terms on the right hand side gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$
But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$
So, substituting into the equation for H gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$
But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure.  So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$