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. `-20.75 xx (373)/(273)kJ`

Answer

A

Explanation

By Kirchhoff.s equation : `(Delta H_(T_(2)) - Delta H_(T_(1)))/(T_(2)- T_(1)) = C_(p)` (Product) `-C_(p)` (Reactant)
`(Delta H_(T_(2)) - (-20.75))/(Delta T)= 0 rArr Delta H_(T_(2))= -20.75kJ`