1.

What is the value of `DeltaS_(surr) ` for the following reaction at 298 K- ` 6CO_(2(g))+ 6 H_(2)O _(1) to C_(6)H_(12)O_(6(s))+6O_(2(g))` Given tahat : ` DeltaG^(@) = 2879 kj "mol"^(-1)` `DeltaS = -210 JK^(-1)"mol"^(-1)`

Answer» Given : `6CO_(2(g))+6H_(2)O_(1) to C_(6)H_(12)O_(6(s)) + 6O_(2(g))`
` DeltaG^(@) = 2879 "KJ mol"^(-1)`,
T =298 K `DeltaH^(@) = ?`
` Delta G^(@) = DeltaH - T DeltaS`
` :. DeltaH = DeltaG^(@) - TDeltaS`
` = 2879 + 298 (-0*210)`
` = 2879 - 62*58`
` = 2816*42 "KJ mol"^(-1)`
Since `DeltaH gt 0` the reaction is endothermic , and system absorbs heat from surroundings . Hence , system loses heat,
` :. Delta H_("surr") = - 2816*42 "KJ mol"^(-1)`
` :. Delta S _("surr") = (DeltaH_("surr"))/T = (-2816*42)/298 = -9*45 "KJK"^(-1)`
` Delta S_("surr") = - 9*45 "KJK^(-1)`
Hence , the value of `DeltaS_("surr")` for the given reaction is ` - 9.45 "KJK"^(-1)`


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