Question

Which of the above given curves is wrongly presented?

• A. $C\to C{O}_2$
• B. $Pb\to PbO$
• C. $Zr\to Zr{O}_2$
• D. $Mg\to MgO$

Answer 1

The correct option is B $Pb\to PbO$

For the transformation $Pb\to PbO$, the curve is wrongly represented.

Consider the reaction $2Pb\left(s\right)+O_2\left(g\right)\to 2PbO\left(s\right)$

This is the combustion reaction. It is exothermic in nature.

Hence, for this reaction, $\Delta H$ is negative.

For the above reaction, $\Delta n<0$. Hence, $\Delta S$ is negative.

At low temperature, $\Delta G$ will be negative, as the temperature increases, $\Delta G$ value increases and becomes positive.