The value of `1og_(10)` K for a reaction `AhArrB` is:
(Given, `Delta_(r)H_(298K)^(@)=-54.07kJ" "mol^(-1),Delta_(r)S_(298K)^(@)=10JK^(-1)" "mol^(-1) and R=8.314JK^(-1)" "mol^(-1),2.303xx8.314xx298=5705`)

Options

A. 5
B. 10
C. 95
D. 100

Answer

B

Explanation

For the equilibrium, `AhArrB`
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`DeltaG^(@)=-2.303RTlog_(10)K`
( `K` is equilibrium constant)
`-2.303RTlog_(10)K=DeltaH^(@)-TDeltaS^(@)`
`2.303RTlog_(10)K=TDeltaS^(@)-DeltaH^(@)`
`log_(10)K=(T DeltaS^(@)-DeltaH^(@))/(2.303RT)`
`=((298xx10+54.07xx1000)/5705)=10`