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Calculate the heat of formation of anhydrous Al_(2)Cl_(6) from the following data : (i) 2Al(s)+6HCl(aq.) to Al_(2)Cl_(6)(aq.)+3H_(2),DeltaH=-244 kcal (ii) H_(2)(g)+Cl_(2)(g) to 2HCl (g),DeltaH=-44 kcal (iii) HCl(g)+aq. to HCl(aq),DeltaH=-17.5 kcal (iv) Al_(2)Cl_(6)(s)+aq. to Al_(2)Cl_(6)(aq.),DeltaH=-153.7 kcal |
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Answer» Solution :we have to calculate `DeltaH` of the equation, (v) `2Al(s)+3Cl_(2)(g) to Al_(2)Cl_(6)(s),DeltaH=?` in the GIVEN equation we see that Equation (iii) does not contain any SPECIES Eqn. (v). Let us thus first consider Eqn. (i), (ii) and (iv) only and apply, `[Eqn. (i)+3xxEqn. (ii)-Eqn.(iv)]`, we get, `2Al(s)+6HCl(aq.)+3H_(2)(g)+3Cl_(2)(g)-Al_(2)Cl_(6)(s)-aq. to` `Al_(2)Cl_(6)(aq.)+3H_(2)(g)+6HCl(g)-Al_(2)Cl_(6)(aq.),` `DeltaH=-244+3xx(-44)-(-153.7)` or `2Al(s)+3Cl_(2)(g)+6HCl(aq)-aq=Al_(2)Cl_(6)(s)+6HCl(g),` `DeltaH=-222.3kcal.` Now MULTIPLYING Eqn. (iii) by `6` and ADDING it to the equation just above we get, `6HCl(g)+aq+2Al(s)+3Cl_(2)(g)+6HCl(aq)-aq to ` `6HCl(aq)+Al_(2)Cl_(6)(s)+6HCl(g),` `DeltaH=6xx(-17.5)+(-222.3)` or `2Al(s)+3Cl_(2)(g) to Al_(2)Cl_(6)(s),DeltaH=-327.3kcal.` |
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