In the Haber 's process of ammonia manufacture:
`N_(2) (g)+3H_(2) (g) to 2NH_(3) (g), triangleH_(298K)^(@)=-92.2kJ`
If `N_(2) (g), H_(2) (g) and NH_(3) (g)" have "C_(p) (JK^(-1) mol^(-1))` values 29.1, 28.8 and 35.1 respectively then if `C_p` is independent of temperature, the reaction at `100^@C` as compared to that of `25^@C` will be

Options

A. More endothermic
B. Less endothermic
C. More exothermic
D. Less exothermic

Answer

C

Explanation

`Delta H^(o)= -92.2`
`DeltaC_(p)=2C_(p)(NH_(3),g)-C_(p)(N_(2),g)-3C_(p)(H_(2))=2xx35.1-29.1-3xx28.8`
`=70.2-29.1-86.4= -45.3 J//K`
`DeltaH_(100^(@)C)^(o)=DeltaH_(25^(@)c)^(o)+DeltaC_(p)(Delta T)= -92.2 -(45.3xx75)/(1000)`
`Delta H_(100^(@)C)^(0)= -92.2-1.3975= -95.6 kJ//"mole"`
Then, reaction at `100^(@)C` as compared to that at `25^(@)C` will be more exhothermic.