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.
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
Correct Answer: 1353.9
Solution and Explanation
The Ellingham diagram plots the standard Gibbs free energy change, \(\Delta G^{\circ}\), of an oxidation reaction against temperature. Reaction I describes the oxidation of solid metal M, and it holds from \(T = 300\) K up to the melting point of M. Reaction II describes the oxidation of liquid metal M, and it holds from the melting point up to \(T = 1800\) K.
Both reactions form the same product, \(MO_2(s)\), from one mole of the metal and one mole of oxygen gas. At the melting point itself, the metal is present in both the solid and the liquid state in equilibrium, so the free energy change of oxidation calculated from either expression must be the same number at that one temperature. This is exactly why the Ellingham line for a metal shows a small change in slope at its melting point, not a jump.
Step 2: Write the two given expressions.
\[ \Delta G^{\circ}_I = -338900 - 15.2\,T\ln T + 247T \text{ J (solid metal, up to melting point)} \]
\[ \Delta G^{\circ}_{II} = -390800 - 15.2\,T\ln T + 285.3T \text{ J (liquid metal, above melting point)} \]
Step 3: Equate the two expressions at the melting point.
Let \(T_m\) be the melting point of the metal. Setting \(\Delta G^{\circ}_I = \Delta G^{\circ}_{II}\) at \(T = T_m\) gives
\[ -338900 - 15.2\,T_m\ln T_m + 247T_m = -390800 - 15.2\,T_m\ln T_m + 285.3T_m \]
The term \(-15.2\,T_m\ln T_m\) has the same value on both sides, so it cancels out completely. This leaves a simple linear equation in \(T_m\):
\[ -338900 + 247T_m = -390800 + 285.3T_m \]
Step 4: Solve the linear equation for the melting point.
Bring the temperature terms to one side and the constants to the other:
\[ 390800 - 338900 = 285.3T_m - 247T_m \]
\[ 51900 = 38.3\,T_m \]
\[ T_m = \frac{51900}{38.3} = 1355.09\ldots\ \text{K} \]
Step 5: Final Answer.
Rounded off to one decimal place, the melting point of the metal M is \(1355.1\) K, which lies inside the accepted range of 1353.9 to 1356.3 K.
\[ \boxed{T_m \approx 1355.1 \text{ K}} \]
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