What will be `DeltaG` for the reaction at `25^(@)C` when partial pressures of reactants `H_(2), CO_(2),H_(2)O and CO` are 10, 20, 0.02 and 0.01 atm respectively. (Given : `G_(H_(2)O)^(@)=-"228.58 kJ, "G_(CO)^(@)=-"137,15 kJ, "G_(CO)^(@)=-"394.37 kJ"`.)A. `+5.61kJ`B. `-5.61 kJ`C. `7.09 kJ`D. `-8.13kJ`

What will be `DeltaG` for the reaction at `25^(@)C` when partial press

1.	What will be `DeltaG` for the reaction at `25^(@)C` when partial pressures of reactants `H_(2), CO_(2),H_(2)O and CO` are 10, 20, 0.02 and 0.01 atm respectively. (Given : `G_(H_(2)O)^(@)=-"228.58 kJ, "G_(CO)^(@)=-"137,15 kJ, "G_(CO)^(@)=-"394.37 kJ"`.)A. `+5.61kJ`B. `-5.61 kJ`C. `7.09 kJ`D. `-8.13kJ`
Answer» Correct Answer - B `H_(2(g))+CO_(2(g))hArr H_(2)O_((g))+CO_((g))` `DeltaG^(@)" for reaction "=[G_(H_(2)O)^(@)+G_(CO)^(@)]-[G_(H_(2))^(@)+G_(CO_(2))^(@)]` `[-228.58-137.15]-[0-394.37]=28.64kJ` Also, `DeltaG=DeltaG^(@)+2.303 RT log Q` `=28.64 +2.303xx8.314 xx10^(-3)xx298log.(0.02xx0.01)/(10xx20)` `Delta=-5.61kJ` Thus, reaction will occur in forward direction.

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