With reference to the Ellingham diagram, the standard Gibbs free energy change for the oxidation of solid metal M(s) and liquid metal M(l) is given below.
Reaction I: \( M(s) + O_2(g) \rightarrow MO_2(s) \)
\[ \Delta G^{\circ} = (-338900 - 15.2\,T\ln T + 247T) \text{ Joules} \]
from \(T = 300\) K to the melting point.
Reaction II: \( M(l) + O_2(g) \rightarrow MO_2(s) \)
\[ \Delta G^{\circ} = (-390800 - 15.2\,T\ln T + 285.3T) \text{ Joules} \]
from the melting point to \(T = 1800\) K.
The melting point of the metal M (rounded off to one decimal place) is _______ Kelvin.
Show Hint
At the metal's melting point, both Gibbs free energy expressions must give the same value. Equate Reaction I and Reaction II at T and solve the resulting linear equation for T.
Step 1: Set up the problem as finding where the two Gibbs energy lines cross.
The two given formulas are two expressions for the same oxidation reaction, one valid below the melting point and one above it. On the Ellingham diagram, a plot of $\Delta G^{\circ}$ against $T$, these two lines meet at exactly one point, the melting point of the metal, because the free energy of the reaction cannot jump suddenly at a single temperature. So instead of comparing the two formulas piece by piece, define a difference function
\[
h(T) = \Delta G^{\circ}_I(T) - \Delta G^{\circ}_{II}(T)
\]
and look for the temperature where $h(T)=0$. Wherever $h(T)=0$, the two lines cross, and that crossing point is the melting point asked for.
Step 2: Build the difference function.
\[
h(T) = \big(-338900 - 15.2\,T\ln T + 247T\big) - \big(-390800 - 15.2\,T\ln T + 285.3T\big)
\]
The $-15.2\,T\ln T$ pieces are identical in both brackets, so they cancel when we subtract. This is a useful check in itself: both reactions share the same gas-phase entropy term, since both consume one mole of $O_2(g)$, so the logarithmic part was always going to drop out of the difference. What is left is only the constant and linear parts:
\[
h(T) = (-338900 + 390800) + (247 - 285.3)T = 51900 - 38.3T
\]
Step 3: Solve $h(T)=0$.
\[
51900 - 38.3T = 0
\]
\[
T = \frac{51900}{38.3}
\]
Carrying out the division, $T = 1355.09\ldots$ K.
Step 4: Round and check against the range.
Rounded to one decimal place, $T = 1355.1$ K.
Step 5: Confirm by substituting back into the original expressions.
As a cross-check, plug $T=1355.1$ K into both original formulas and make sure they give the same $\Delta G^{\circ}$. Both share the term $-15.2\,T\ln T$, which evaluates to the same number in each, so it is enough to compare the remaining linear parts:
\[
-338900+247(1355.1) = -338900+334709.7=-4190.3
\]
\[
-390800+285.3(1355.1) = -390800+386610.03=-4189.97
\]
The two values agree to well within rounding error, which confirms $T=1355.1$ K is the crossing point.
Step 6: State the final answer.
This value falls comfortably inside the window of 1353.9 K to 1356.3 K that the answer key accepts, so the melting point checks out.
\[
\boxed{T \approx 1355.1 \text{ K}}
\]
