Question

In Ellingham diagram, to indicate the boiling point of oxide of a metal and that of a metal, the nature of a curve will have turning point with

• A. Increasing slope and decreasing slope respectively.
• B. Decreasing slope and increasing slope respectively.
• C. Increasing and increasing slope respectively.
• D. Decreasing and decreasing slope respectively.

Answer 1

Answer: (B)

Decreasing slope and increasing slope respectively.

Answer 2

Ellingham diagrams plot the standard Gibbs free energy change ($\Delta G^\circ$) for the formation of metal oxides as a function of temperature (T). The general reaction is $xM(s) + y/2 O_2(g) \rightarrow M_xO_y(s)$. The relationship between $\Delta G^\circ$ and T is given by the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

In an Ellingham diagram, the plot of $\Delta G^\circ$ versus T is approximately a straight line with slope equal to $-\Delta S^\circ$. The standard entropy change ($\Delta S^\circ$) for the formation of a solid metal oxide from a solid metal and gaseous oxygen is generally negative because the number of moles of gas decreases (oxygen is consumed). Thus, the slope $-\Delta S^\circ$ is generally positive.

Phase transitions of the reactants or products cause abrupt changes in the value of $\Delta S^\circ$, and hence in the slope of the $\Delta G^\circ$ vs T curve.

1. Boiling point of the metal oxide ($M_xO_y$): At the boiling point of the oxide, the oxide changes state from solid to gas (or liquid to gas). Let's consider the transition from solid to gas for simplicity, as the entropy change is most significant here. The reaction changes from $xM(s) + y/2 O_2(g) \rightarrow M_xO_y(s)$ to $xM(s) + y/2 O_2(g) \rightarrow M_xO_y(g)$. The original entropy change is $\Delta S^\circ_{solid\_oxide} = S^\circ(M_xO_y(s)) - xS^\circ(M(s)) - y/2 S^\circ(O_2(g))$. The new entropy change is $\Delta S^\circ_{gaseous\_oxide} = S^\circ(M_xO_y(g)) - xS^\circ(M(s)) - y/2 S^\circ(O_2(g))$. Since $S^\circ(M_xO_y(g)) > S^\circ(M_xO_y(s))$, we have $\Delta S^\circ_{gaseous\_oxide} > \Delta S^\circ_{solid\_oxide}$. The slope of the curve is $-\Delta S^\circ$. The new slope is $-\Delta S^\circ_{gaseous\_oxide}$ and the old slope is $-\Delta S^\circ_{solid\_oxide}$. Since $\Delta S^\circ_{gaseous\_oxide} > \Delta S^\circ_{solid\_oxide}$, it follows that $-\Delta S^\circ_{gaseous\_oxide} < -\Delta S^\circ_{solid\_oxide}$. Thus, at the boiling point of the oxide, the slope of the curve decreases.

2. Boiling point of the metal ($M$): At the boiling point of the metal, the metal changes state from solid to gas (or liquid to gas). Let's consider the transition from solid to gas. The reaction changes from $xM(s) + y/2 O_2(g) \rightarrow M_xO_y(s)$ to $xM(g) + y/2 O_2(g) \rightarrow M_xO_y(s)$. The original entropy change is $\Delta S^\circ_{solid\_metal} = S^\circ(M_xO_y(s)) - xS^\circ(M(s)) - y/2 S^\circ(O_2(g))$. The new entropy change is $\Delta S^\circ_{gaseous\_metal} = S^\circ(M_xO_y(s)) - xS^\circ(M(g)) - y/2 S^\circ(O_2(g))$. Since $S^\circ(M(g)) > S^\circ(M(s))$, we have $-xS^\circ(M(g)) < -xS^\circ(M(s))$. Therefore, $\Delta S^\circ_{gaseous\_metal} < \Delta S^\circ_{solid\_metal}$. The slope of the curve is $-\Delta S^\circ$. The new slope is $-\Delta S^\circ_{gaseous\_metal}$ and the old slope is $-\Delta S^\circ_{solid\_metal}$. Since $\Delta S^\circ_{gaseous\_metal} < \Delta S^\circ_{solid\_metal}$, it follows that $-\Delta S^\circ_{gaseous\_metal} > -\Delta S^\circ_{solid\_metal}$. Thus, at the boiling point of the metal, the slope of the curve increases.

The question asks about the nature of the curve at the turning point corresponding to the boiling point of the oxide and that of the metal, respectively. Based on our analysis, the slope of the curve decreases at the boiling point of the oxide and increases at the boiling point of the metal.