The value of enthalpy change, H_β − H_α (in J mol⁻¹), at 300 K is:

The correct answer is 300.

From the given image, we observe a graph plotting $S_T-S_0$ (in J mol⁻¹ K⁻¹) on the y-axis against Temperature (in K) on the x-axis for two phases, α and β.

Notice that the lines for α and β are straight lines passing through the origin (or having constant slope), which indicates that:

$S_T-S_0=\int \frac{C_p}{T}dT=C_p\ln \left(T\right)$ if C_p depends on temperature, or for constant slopes in such temperature plots where $S_T-S_0=\frac{C_p}{T}·T=C_p$ or linear dependence. Specifically, at T = 600 K:

For phase β:
$(S_T-S_0{)}_{\beta }=6\text{J mol}{}^{-1}\text{K}{}^{-1}$

For phase α:
$(S_T-S_0{)}_{\alpha }=5\text{J mol}{}^{-1}\text{K}{}^{-1}$

The slope of each straight line gives the value of $\frac{S_T-S_0}{T}$. Thus, at T = 300 K, the value of $S_T-S_0$ scales proportionally with temperature:

$\Delta C_p=(S_T-S_0{)}_{\beta }-(S_T-S_0{)}_{\alpha }=6-5=1\text{J mol}{}^{-1}\text{K}{}^{-1}$

Using the relation for enthalpy change $\Delta ;H=H_{\beta }-H_{\alpha }=T·\Delta S$ at T = 300 K:

$H_{\beta }-H_{\alpha }=300\text{K}\times 1\text{J mol}{}^{-1}\text{K}{}^{-1}=300\text{J mol}{}^{-1}$

Therefore, the value of the enthalpy change $H_{\beta }-H_{\alpha }$ at 300 K is 300 J mol⁻¹.