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Calculate the amount of heat evolved when (i) 500 cm^(3) of 0.1 Mhydrochloric acid is mixed with 200 cm^(3) of 0.2 M sodium hydroxide solution (ii) 200 cm^(3) of0.2 M sulphuric acid is mixed with 400 cm^(3) of 0.5 M potassium hydroxide solution. Assuming that the specific heat of water is 4.18 J K^(-1) g^(-1) and ignoring the heat absorbed by the container, thermometer, stirrer etc, what would be the rise in temperature in each of the above cases ? |
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Answer» Solution :(i) `500 cm^(3)` of 0.1M HCl`= (0.1)/(1000) XX 500`MOLE of `HCl = 0.05` mole of HCl `= 0.05` mole of `H^(+)` ions ` 200 cm^(3)` of0.2 M NaOH `= (0.2)/( 1000) xx 200 `mole of NaOH `=0.04` mole of NaOH `=0.04` mole of `OH^(-)`ions Thus, `0.04` mole of `H^(+)`ions will combinewith `0.04` mole of `OH^(-1) ` ions to FORM moleof `H_(2)O` and 0.01 mole of `H^(+)` ions will remain unreacted. Heart evolved when 1 mole of `H^(+)`ions combine with 1 mole of `OH^(_)` ions `= 57.1 kJ`. `:. ` Heat evolved when 0.04 mole of `H^(+)`ions combine with0.04 mole of `OH^(-)` ions `= 57.1xx 0.04 = 2.284 kJ` (ii) `200 cm^(3)` of 0.2M `H_(2)SO_(4) = (0.2)/(1000) xx 200` mole of `H_(2)SO_(4) = 0.04` mole of `H_(2)SO_(4) = 0.08` mole of `H^(+)` ions ` 400 cm^(3)` of 0.5 M KOH `= (0.5)/(1000) xx 400` mole o KOH `= 0.2` mole of `OH^(-)` ions Thus, 0.08 mole of `H^(+)` ions will neutralize`0.08` mole of `OH^(-)` iosn. (out of 0.2 mole of `OH^(-)` ions) to form `0.08` mole of `H_(2) O` Hence, heat evolved `= 57.1 xx 0.08 = 4.568kJ` In case (i) , heat produced `= 2.284 kJ = 2284J` Total mass of the solution `= 500 + 200 = 700 g` Specific heat `= 4.18 J K^(-1) g^(-1)` `Q = m xx C xx Delta t ``:.Delta t = (Q)/(m xx C) = ( 2284)/( 700 xx 4.18 ) = 0.78 ^(@)` In case (ii), heat produced `= 4.568 kJ = 4568 J ` Total mass of the solution ` = 200 + 400 = 600 g ``:. Delta t = (Q) /( m xx C ) = ( 4568) /( 600 xx 4.18) = 1.82 ^(@)` |
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