The enthalpy of certain reaction at `273 K` is `-20.75 kJ`. The enthalpy of the same reaction at `373 K` provided heat capacities fo reactants and products are the same) will be

Options

A. `-20.75 kJ`
B. `-2075 kJ`
C. zero
D. `-20.75xx(373)/(273)kJ`

Answer

A

Explanation

By Kirchoff's equation `(DeltaH_(2)-DeltaH_(1))/(T_(2)-T_(1))=DeltaC_(P)`
Since `(C_(P))` reactant `=(C_(P))` Products, `DeltaC_(P)=0`
Hence `DeltaH_(2)-DeltaH_(1)=0`
or `DeltaH_(2)=DeltaH_(1)=-20.75 kJ`