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The heats of combustion of ammonia and hydrogen are 9.06 and 68.9 kcal respectively. Calculate the heat of formation of ammonia. |
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Answer» Solution :Given that, (i) `{:(NH_(3)(g)+(3)/(4)O_(2)(g) to (1)/(2)N_(2)(g)+(3)/(2)H_(2)O,,DeltaH=-9.06 "kcal"):}` (ii) `{:(H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),,DeltaH=-68.9 "kcal"):}` ( The negative signs are taken as the combustion process is exothermic) we have to calculate `DeltaH` of the FOLLOWING reaction : (iii) `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3)(g),DeltaH=?` Since `H_(2)` in Equation (iii) and in Equation (ii) is on ethe same side and `NH_(3)` in Equation (iii) and in Equation (i) is on OPPOSITE sides, we FIRST multiply Equation (ii) by (3)/(2) to equate number of MOLES of `H_(2)` in equations (ii) and (iii) . Then SUBTRACTING Equation (i) from equation (ii) `{i.e.,(3)/(2)xx"Equation"(ii)-"Equation"(i)},"we get",` `(3)/(2)H_(2)(g)+(3)/(4)O_(2)(g)-NH_(3)(g)-(3)/(4)O_(2)(g) to` `(3)/(2)H_(2)O(g)-(1)/(2)N_(2)(g)-(3)/(2)H_(2)O(g),` `DeltaH=(3)/(2)xx(-68.9)-(-9.06)` or `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3),DeltaH=-94.29 "kcal"` |
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