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The standard enthalpy of formation of `C_(2)H_(4)(l)`, `CO_(2)(g)` and `H_(2)O(l)` are `52`, `394` and `-286 kJ mol^(-1)` respectively. Then the amount of heat evolved by burning `7g` of `C_(2)H_(4)(g)` isA. `1412 kJ`B. `9884 kJ`C. `353 kJ`D. `706 kJ` |
Answer» Correct Answer - C Desired equation `C_(2)H_(4)(g)+3O_(2)to2CO_(2)+2H_(2)O` , `DeltaH=?` `DeltaH=[2DeltaH_(f)^(@)(CO_(2))+2DeltaH_(f)^(@)(H_(2)O)-DeltaH_(f)^(@)(C_(2)H_(2))-3DeltaH_(f)^(@)(O_(2))]` `=2(-394)+2(-286)-52-0=1412 kJ mol^(-1)` or `1` mol (287) of `C_(2)H_(4)` will give heat `=1412 kJ` `:. 7 g` of `C_(2)H_(4)` will give `=(1412)/(28)xx7=353 kJ` |
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