Standard Gibbs free energies of formation of some solid oxides per mole of O\(_2\) at 1000 K are given below.
\[
{SiO}_2: -728 \, {kJ}, \, {TiO}_2: -737 \, {kJ}, \, {VO}: -712 \, {kJ}, \, {MnO}: -624 \, {kJ}
\]
Regarding thermodynamic feasibility of oxide reduction, which of the following statements is/are CORRECT under standard conditions at 1000 K?
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- Si can reduce TiO\(_2\)
- Mn can reduce VO
- Ti can reduce MnO
- V can reduce SiO\(_2\)
The Correct Option is C
Solution and Explanation
In thermodynamics, the feasibility of a reaction is based on the Gibbs free energy change (\(\Delta G\)). A reduction reaction is thermodynamically favorable if the reduction of the metal oxide leads to a positive \(\Delta G\), indicating that the reaction will proceed spontaneously. The substance with the more negative \(\Delta G\) is more likely to be reduced. Let's evaluate each option based on the Gibbs free energy values provided:
Step 1: Analysis of each option
- Option (A): "Si can reduce TiO\(_2\)" - Incorrect: To reduce TiO\(_2\), Si would need a more negative Gibbs free energy than TiO\(_2\). Since the Gibbs free energy of formation for TiO\(_2\) is \(-737 \, {kJ}\) and Si has a more positive Gibbs free energy for its oxide formation, Si cannot reduce TiO\(_2\) under standard conditions.
- Option (B): "Mn can reduce VO" - Incorrect: To reduce VO, Mn would need to have a more negative Gibbs free energy than VO. The Gibbs free energy for VO is \(-712 \, {kJ}\), and Mn has a less negative value, meaning Mn cannot reduce VO.
- Option (C): "Ti can reduce MnO" - Correct: To reduce MnO, Ti would need to have a more negative Gibbs free energy. The Gibbs free energy for MnO is \(-624 \, {kJ}\), while Ti has a more negative Gibbs free energy of \(-737 \, {kJ}\), which means Ti can reduce MnO under standard conditions. This makes Option C correct.
- Option (D): "V can reduce SiO\(_2\)" - Incorrect: To reduce SiO\(_2\), V would need a more negative Gibbs free energy than SiO\(_2\). Since the Gibbs free energy for SiO\(_2\) is \(-728 \, {kJ}\), and V has a less negative value, V cannot reduce SiO\(_2\).
Step 2: Conclusion The correct answer is Option C. Ti can reduce MnO because its Gibbs free energy of formation is more negative than that of MnO.
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