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Given the reaction at 967^(@)C and 1 atm...

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

A

`156.6KJ`

B

`165.6KJ`

C

`16.6KJ`

D

`1.656KJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaH=DeltaE+Deltan_(g)RT`
Here `Deltan_(g)=1,T=967+273=1240K`
`176=DeltaE+((1xx8.314xx1240)/(1000))KJ`
`176=DeltaE+10.309KJ`
`DeltaE=176-10.309KJ`
`=165.69KJ`
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