According to the following diagram, A reduces BO2 when the temperature is (a) 1200°C but

Q: Why do the lines in the Ellingham diagram slope upwards?

The formation of a solid oxide from a solid metal and oxygen gas consumes gas, leading to a decrease in randomness or entropy ($\Delta S < 0$). Since $\Delta G = \Delta H - T\Delta S$, a negative $\Delta S$ makes the $-T\Delta S$ term positive, causing the overall $\Delta G$ to increase (become less negative) as temperature rises.

Q: What exactly happens at the intersection point of $1400^\circ\text{C}$?

At this temperature, $\Delta G^\circ$ for both oxide formations is exactly equal. The coupled reaction $A + BO_2 \rightleftharpoons AO_2 + B$ is at perfect equilibrium, meaning the overall $\Delta G^\circ_{\text{reaction}} = 0$.

Q: Can $B$ reduce $AO_2$ at $1500^\circ\text{C}$?

No. At $1500^\circ\text{C}$, the curve for $A$ is lower than $B$, meaning $AO_2$ is more stable than $BO_2$. Thus, $A$ will reduce $BO_2$, not the other way around.

Q: Why is the y-axis labeled with negative values at the top?

Actually, the y-axis has zero at the top and becomes more negative as you go down. A lower position on the graph means a more negative $\Delta G^\circ$, indicating greater thermodynamic stability of the oxide.

Q: Does this diagram tell us how fast the reduction will happen?

No, the Ellingham diagram only provides thermodynamic information (feasibility). It tells us *if* a reaction can happen spontaneously, but it gives absolutely no information about *how fast* (kinetics) the reaction will occur.

Question

According to the following diagram, $A$ reduces $BO_2$ when the temperature is Options: (A) $> 1200^\circ ext{C}$ but $< 1400^\circ ext{C}$ (B) $>1400^\circ ext{C}$ (C) $<1200^\circ ext{C}$ (D) $<1400^\circ ext{C}$

Sigma Percentile · Accepted Answer

### The Battlefield: Gibbs Free Energy vs. Temperature

Welcome to the fascinating world of metallurgy, where elements constantly battle for oxygen! The Ellingham diagram is our ultimate map of this battlefield. It plots the standard Gibbs free energy of formation ($\Delta G^\circ$) of various metal oxides against temperature ($T$).

The y-axis is a bit counter-intuitive at first glance: the values become more negative as you go *down*. In thermodynamics, a more negative $\Delta G^\circ$ means a reaction is highly spontaneous and the resulting product is incredibly stable. Therefore, the lower a metal's curve is on this diagram, the more stable its oxide is compared to the ones above it.

### The Golden Rule of Reduction

How do we use this diagram to choose a reducing agent? The rule is beautifully simple: **A metal whose curve lies lower on the Ellingham diagram can reduce the oxide of any metal whose curve lies above it.**

Why? Because the metal lower down has a stronger affinity for oxygen. If you mix them together, the lower metal will literally rip the oxygen atoms away from the higher metal's oxide. This is the core principle behind extracting pure metals from their ores.

### Analyzing the Contenders: Metal A vs. Metal B

Let's look at the specific curves given in our problem. We have two competing reactions:
1. $A + O_2 ightarrow AO_2$ (Blue Line)
2. $B + O_2 ightarrow BO_2$ (Red Line)

If we look at the left side of the graph (temperatures below $1400^\circ	ext{C}$), the curve for metal $B$ is clearly below the curve for metal $A$. This means that in this cooler region, $BO_2$ is more stable than $AO_2$. Consequently, metal $B$ acts as the boss here—it can easily reduce $AO_2$ to pure metal $A$.

### The Turning Point: $1400^\circ	ext{C}$

But thermodynamics is dynamic! As we heat things up, the slopes of the lines dictate a change in destiny. Notice the exact point where the two lines intersect: $T = 1400^\circ	ext{C}$.

At this precise temperature, $\Delta G^\circ_{AO_2} = \Delta G^\circ_{BO_2}$. The stabilities of both oxides are perfectly matched. If you were to mix $A$, $B$, $AO_2$, and $BO_2$ at this temperature, they would sit in a state of chemical equilibrium. Neither metal has the upper hand.

### The Final Verdict

The question asks us to find the condition where **$A$ reduces $BO_2$**. For this to happen, metal $A$ must be the stronger reducing agent, which means its curve must be *lower* than $B$'s curve.

Looking past the intersection point, into the region where $T > 1400^\circ	ext{C}$, we see exactly that! The blue line for $A$ dips below the red line for $B$. In this high-temperature zone, $AO_2$ becomes the more stable oxide.

Therefore, if we heat the mixture above $1400^\circ	ext{C}$, metal $A$ will aggressively steal oxygen from $BO_2$, leaving behind pure metal $B$. The correct condition is **$> 1400^\circ	ext{C}$**.

Animated Solution for Chemistry - Principles of Metallurgy and Extraction: According to the following diagram, $A$ reduces $B O_{2}$ when the temperature is

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Visualized Solution

\text{The Ellingham Diagram}

\text{Thermodynamic Principle of Reduction}

\text{Analyzing the Curves}

\text{The Intersection Point}

\text{Below } 1400^\circ\text{C}

\text{Above } 1400^\circ\text{C}

\text{Coupled Reactions}

The Sigma Insight: Principles of Metallurgy and Extraction

The Battlefield

The Golden Rule of Reduction

Analyzing the Contenders

The Turning Point: $140 0^{\circ} C$

The Final Verdict

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Animated Solution for Chemistry - Principles of Metallurgy and Extraction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } According to the following diagram, A reduces BO2​ when the temperature is