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Ethylene on combustion gives carbon dioxide and water. Its heat of combustion is `1410.0 kJ mol^(-1)`. If the heat of formation of `CO_(2)` and `H_(2)O` are `393.3 kJ` and `286.2 kJ`, respectively. Calculate the heat of formation of ethylene. |
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Answer» a. `CH_(2) = CH_(2)(g) +3O_(2)(g) rarr 2CO_(2)(g) +2H_(2)O DeltaH_(1) - 1410 K mol^(-1)` b. `C(s) +O_(2)(g) rarr CO_(2)(g)` `DeltaH_(2) = - 393.3 kJ mol^(-1)` c. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l)` `DeltaH_(2) =- 286.2 kJ mol^(-1)` To calculate `DeltaH` of the reaction: `2C(s) +2H_(2)(g) rarr CH_(2) = CH_(2)(g) DeltaH = ?` `:. DeltaH = 2DeltaH_(2) +2DeltaH_(2) - DeltaH_(1)` `= 2 xx -393.3 -2 xx 286.2 -(-1410)` `= 51.0 kJ mol^(-1)` |
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