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Calculate the magnitude of standard entr...

Calculate the magnitude of standard entropy change for reaction `X hArr Y` if `DeltaH^(@) = 25 KJ` and `K_("eq") "is" 10^(-7)` at `300` K.

Text Solution

Verified by Experts

The correct Answer is:
51
Given : `DeltaH^(@) = 25KJ`
`T=300 k`
`K_("eq") = 10^(-7)`
Asked : standard entropy change
Formula :`DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
Since `DeltaG^(@) =- 2.303 "RT log K"_("eq")`
`therefore" "-2.303 "RT log K"_("eq") = DeltaH^(@)-TDeltaS^(@)`
Explanation: `DeltaG^(@)`= Standard free energy change
`DeltaH^(@) ` = Standard enthalpy change
T=Temperature
`DeltaS^(@)` = standard enthropy change
R = Gas constant
`K_("eq")` = equilibrium constant
Substitution & calculation:
`-2.303 xx (25//3) xx (300) "log"10^(-7)=25000-(300) DeltaS^(@)`
`DeltaS^(@) = -51 J//K.`
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