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The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :A. `+46.0 kJ`B. `+92.0 kJ`C. `-23 kJ`D. `-92 kJ` |
Answer» Correct Answer - B The given reaction involve decomposition of `2` mol of ammonia `DeltaH_(f)(NH_(3))=-46.0`, `DeltaH_("decomposition")"of" NH_(3)=+46` `:. DeltaH "of given reaction"=2xx46=92 kJ` |
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