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The value of Delta G^@ for formation of Cr_2O_3 is -540 kJ mol^(-1) and that of Al_2O_3 is -827 kJ mol^(-1). Is the reduction of Cr_2O_3 possible with Al ? |
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Answer» Solution :`4/3 Al_((s)) + O_(2(g)) to 2/3 Al_2O_(3(s)) , Delta_f G^(@) = -827 kJ "".....(i)` `4/3 Cr_((s)) + O_(2(g)) to 2/3 Cr_(2)O_(3(s)) , Delta_f G^@ = -540kJ "" ....(II)` Subtracting EQUATION (ii) from (i) we get, `4/3 Al_((s)) + 2/3 Cr_(2)O_(3(s)) to 2/3 Al_2O_(3(s)) + 4/3Cr_((s))` `Delta_f G^@ = -287 kJ mol^(-1) "" .....(iii)` SINCE net `Delta_fG^@` is negative, the reaction is thermodynamically feasible and therefore AL can reduce `Cr_2O_3`. |
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