chemistry-class-12-unit-06-chapter-02-isolation-of-metals-l-2-3-2r4tfvp5als

<h4 id="successextraction-of-crude-metal-from-concentrated-oresuccess-1"><Success>Extraction of crude metal from concentrated ore</Success></h4>
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<p><strong>Important question</strong></p>
<p><strong>Question</strong> - Which is the reducing agent , suitable for an oxide (hint - thermodynamics may provide answer) ?</p>
<p><strong>Answer</strong> - In reduction of metal we want following steps</p>
<ul>
<li>ZnO → Zn +1/2 O₂  (endothermic), $\Delta S$ is positive… (1)</li>
<li>It is reverse of Zn + ½ O₂ → ZnO (exothermic) $\Delta S$ is negative… (2)</li>
<li>As we know gibbs energy , $\Delta G$ for any process at any specified temperature is</li>
</ul>
<p>$\Delta G=\Delta H-T \Delta S$ & $\Delta G \degree = -RTInK$</p>
<ul>
<li>For spontaneous process $\Delta G$ should be (-ve)</li>
</ul>
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<h4 id="successextraction-of-crude-metal-from-concentrated-oresuccess-2"><Success>Extraction of crude metal from concentrated ore</Success></h4>
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<p><strong>Answer</strong></p>
<ul>
<li>For $\Delta G$ is negative $\rarr +\Delta H < - T\Delta S /$  K has to be positive</li>
<li>On increasing T the value of $T\Delta S$ would increase and ultimately $\Delta G$ - ve</li>
<li>This is impractical as very high T would be needed</li>
<li>If net $\Delta G$ of one with other reaction is - ve the overall reaction will occur</li>
<li>So in reduction of metal oxide we couple two reactions so that sum of their $\Delta G$ is negative and of high magnitude</li>
</ul>
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<h4 id="successextraction-of-crude-metal-from-concentrated-oresuccess-3"><Success>Extraction of crude metal from concentrated ore</Success></h4>
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<p><strong>Answer (3/3)</strong></p>
<p>${M}_xO(s) \rarr xM(s/l) + 1/2{O}_2(g)\quad \Delta S$ is +ve</p>
<p>$yM^* + x/2O_2 \rarr M^*yOx \quad (\Delta S$ is -ve)…(1)</p>
<p>$C(S)+1/2O_2(g) \rarr CO(g)$ $\quad$ $\Delta S$ is +ve for 2…(2)</p>
<p>$CO(g)+1/2O_2(g) \rarr CO_2(g)$ $\quad$ $\Delta S$is -ve for 3…(3)</p>
<p>$\Delta S$ first reaction is (+ve)</p>
<ul>
<li>To make $\Delta G$(-ve) , T has to high as it is endothermic reaction</li>
<li>Its combination with any of next three exothermic reactions gives overall ( - ve) $\Delta G$</li>
<li>$C$ or $M^*$ is suitable for reduction</li>
</ul>
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<h3 id="successellingham-diagramsuccess"><Success>Ellingham diagram</Success></h3>
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<p><strong>Important points of ellingham diagram</strong></p>
<ul>
<li>It is the plot of $\Delta G$ vs T</li>
<li>It is straight line and its slope is $\Delta S$</li>
</ul>
<p>$2Zn+O_2 \rarr 2ZnO$</p>
<p>$\Delta G=\Delta H-T\Delta S$</p>
<ul>
<li>It only indicates whether a reaction is possible or not</li>
<li>It does not say about the kinetics of the reduction process i.e how fast it could be</li>
</ul>
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<h3 id="successextraction-of-metal-from-oresuccess"><Success>Extraction of metal from ore</Success></h3>
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<p><strong>Extraction of copper from cuprous oxide[copper(l) oxide]</strong></p>
<ul>
<li>In ellingham diagram $\Delta G\degree$ vs T the $Cu_2O$ line the top region</li>
<li>Lines of C , CO and C , $CO_2$ are at much lower positions in 500-600K region</li>
<li>Oxide ores of copper easy to reduce directly to the by heating with coke</li>
<li>Most of the ores are sulphide and some may also contain iron</li>
<li>The sulphide ores are roasted/smetted to give oxides</li>
</ul>
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<h3 id="successreverberatory-furnacesuccess"><Success>Reverberatory furnace</Success></h3>
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<p>Actual process</p>
<ul>
<li>
<p>Ore is mixed with silica and heated in a reverberatory</p>
</li>
<li>
<p>In the furnace , iron oxide ‘slag off’ as iron silicate</p>
<p>$FeO+SiO_2 \rarr FeSiO(Slag)$</p>
</li>
<li>
<p>Copper is produced in the form of copper matte</p>
</li>
<li>
<p>This contains $Cu_2S$ and FeS</p>
</li>
</ul>
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<h3 id="successextraction-of-metal-from-oresuccess-1"><Success>Extraction of metal from ore</Success></h3>
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<p><strong>Extraction of copper from cuprous oxide[copper(l) oxide]</strong></p>
<ul>
<li>Copper matte is then charged into silica lined convertor</li>
</ul>
<ul>
<li>
<p>Some silica is added and hot air blast is blown</p>
<p>$2FeS+3O_2 \rarr 2FeO+2SO_2$</p>
<p>$FeO+SiO_2 \rarr FeSiO_3$</p>
<p>$2Cu_2S+3O_2 \rarr 2Cu_2O+2SO_2$</p>
<p>$2Cu_2O+Cu_2S \rarr 6Cu+SO_2$</p>
</li>
<li>
<p>Copper is called blister copper</p>
</li>
<li>
<p>It has blistered appearance due to the evolution of $SO_2$</p>
</li>
</ul>
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### <Success>Extraction of metal from ore</Success>
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<p>$Cr_2O_3+Al \rarr Al_2O_3+2Cr \hspace{5 mm} \Delta G\degree =-421 KJ$</p>
<ul>
<li>Thermodynamically feasible at room temp</li>
<li>But does not take place</li>
<li>The change in gibbs energy is related to the equil const K</li>
<li>$\Delta G\degree = -RTInK$</li>
<li>At room temperature , all reactants and products of the given reaction are in the solied state</li>
<li>No equilibrium exists between the reactants and the products</li>
<li>Hence , the reaction does not take place at room temperature</li>
</ul>
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