For a reaction `DeltaC_(p)=2.0+0.2T cal^(@)C` and enthalpy of reaction at `10K ` is `-14.3kcal.` The enthalpy of this reaction at `100K` is `:`

Options

A. `-16.3kcal`
B. `-17.02kcal`
C. `-13.13kcal`
D. `-14.08kcal`

Answer

C

Explanation

`DeltaH_(2)=DeltaH_(1)+int_(10)^(100)DeltaC_(p).dt`
`=-14.3+int_(10)^(100)[2.0+0.2T].dt`
`=-14.3xx10^(3)`
`+[2.0xx(100-10)+0.2xx((100^(2))/(2)-(10^(2))/(2))]`
`=-14.3xx10^(3)+(180+1000-10)`
`=-14.3xx10^(3)+1170=-13.13kcal`