## Answers

Solution cyme B.lag) A(ags Tempezuture 7.190 kJ, mol 3190 3.mot AG 298.15K In dhis questran, we use enuction RT Lne AG 440 A eauilileim e Kea So eauatran ccomus RT dn kea 2.303 RT Jog kear AGO So, From cauation cu - 7.190 x LO mol =-2.303 x %.314 kmot x 298 15k X \og kear log kea I.26 Keas kear Lo1-2 ILib 2-303 RT 310.15 K AG Tenbezeture Conc. o y 17M A4o 37.0°c Conc ofB O.75M O.75 M 1-7 M O.4412 (reaction quoticact) So, FTOm eautton Ci 2.303 x 8.3l4 k.mol X3l015K mol X log (a.uu2).

A4= +l-2,110.

.
33 mol F190 mel -9300. 33 mo 9.30033 KI mel

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