Question

According to the following diagram, $\mathrm{A}$ reduces ${\mathrm{BO}}_2$ when the temperature is:

• A. $>1400°\mathrm{C}$
• B. $<1400°\mathrm{C}$
• C. $>1200°\mathrm{C}$
• D. $<1200°\mathrm{C}$

Answer 1

The correct option is A

$>1400°\mathrm{C}$

Answer: (A)

The explanation for correct option:

In Ellingham’s diagram, the line of the element that lies below can reduce the oxide of the element which lies above it. Therefore, $\mathrm{A}$ to reduce${\mathrm{BO}}_2$, the temperature when the element line $\mathrm{A}$ is below that of${\mathrm{BO}}_2$, according to the graph is when$\mathrm{T}>1400°\mathrm{C}$.

For $\mathrm{T}>1400°\mathrm{C},$$∆{\mathrm{G}}_{\mathrm{r}}<0,\mathrm{for}\mathrm{A}+{\mathrm{BO}}_2\to \mathrm{B}+{\mathrm{AO}}_2$

Hence, option A is correct.