1.

What will be the standard internal energy change for the reaction at 298 K ? `OF_(2(g))+H_(2)O_((g))+2HF_((g)),DeltaH^(@)=-310kJ`A. `-312.5Kj`B. `-125.03kJ`C. `-310 kJ`D. `-156kJ`

Answer» Correct Answer - A
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaH=-310xx10^(3)J, Deltan_(g)=3-2=1, R="8.314 J K"^(-1)"mol"^(-1),`
T = 298 K
`DeltaU=-310 xx10^(3)-(1xx 8.314xx298)`
`=-312.477xx106(3)J=-312.5kJ`


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