According to the following diagram, A reduces BO₂ when the temperature is :

Solution

Chemistry ma’am · 4 steps

The metal whose oxide line sits lower can pull oxygen away from the other, so find where A dips below B.

1. Write the reduction. A reducing BO₂ means A + BO₂ → AO₂ + B. This is the A line reaction minus the B line reaction, so its Δ G° is the difference between the two lines.

   Δ G°_(reduction) = Δ G°(AO₂) - Δ G°(BO₂)

2. Read the graph. This is negative only when the A line lies below the B line. Up to about 1400 °C the B line is lower. For example, at 1000 °C you get roughly -1047 - (-1150) = +103 kJ mol⁻¹, which is not feasible. The B line turns upward near 1250 °C and crosses the A line at about 1400 °C.
3. Check above 1400 °C. At 1600 °C the value is roughly -1015 - (-860) = -155 kJ mol⁻¹. It is negative, so the reduction is feasible.
4. So. A reduces BO₂ only above the crossing point, that is, at temperatures above 1400 °C. This is option (a).

The idea to keep. On an Ellingham diagram, a metal reduces another metal's oxide only where its own oxide line lies lower. The crossing point of the two lines is the threshold temperature.

Answer from the NTA final answer key. Solution by PrepWiser.