Given the reaction at `967^(@)C` and `1 atm`.
`CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)`
`DeltaH=176 kJ mol^(-1)`, then `DeltaE` equals

Options

A. 160 kJ
B. 165.6 kJ
C. 186.4 kJ
D. 180 kJ

Answer

B

Explanation

`CaCO_(3(s)) hArr CaO_((s))+CO_(2(g)) Delta H =176 kJ`
`Delta H = Delta U + Delta nRT therefore Delta U = Delta H - Delta nRT`
`Delta n=(0+1)-0=1`
`Delta therefore Delta U =176 -1xx8.314xx10^(-3)xx1240`
`=176-10.31=165.69 kJ`
`Delta U =165.6 kJ`