The activation energy of most of the reactions occurring at `25^(@)C` is generally `50"kJ.mol"^(-1)` . Calculate the temperature coefficient of such reactions.

Explanation

The temperature coefficient `=(k_(35^(@)C))/(k_(25^(@)C)),`
where, `k_(25^(@)C)and k_(35^(@)C)` are the rate constants at `25^(@)C and 35^(@)C` respectively . We know that ,
`log.(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)-T_(2))),(T_(2)gtT_(1))`
`therefore" For "(k_(2))/(k_(1))=(k_(35^(@)C))/(k_(25^(@)C)), T_(1)=(273+25)=298K, `
`T_(2)=(273+35)=380K andE_(a)=50xx10^(3)"J.mol"^(-1)`
`therefore log.(k_(35^(@)C))/(k_(25^(@)C))=(50xx10^(3))/(2.303xx8.314)((308-298)/(298xx308))`
or , `log.(k_(35^(@)C))/(k_(25^(@)C))=0.2845 therefore(k_(35^(@)C))/(k_(25^(@)C))=1.925`
`therefore` The temperature coefficient of the reaction `=1.925~=2`