InterviewSolution
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InterviewSolution
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For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`. Which relation is correct?A. `2.303 log_(10)K= -(DeltaH^(@))/(RT)+(DeltaS^(@))/(R )`B. `DeltaG=DeltaG^(@)+2.303RT log_(10)Q`C. `K=Ae^(-DeltaH^(@)//RT)`D. all are correct |
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Answer» All are correct. `DeltaG^(@)=-2.303Rtlog_(10)K_(c)` (At eqm. `DeltaG=0`) `DeltaH^(@)-TDeltaS^(@)=-2.303RT log_(10)K` `2.303 log_(10)K_(c)=(DeltaH^(@))/(RT)-(DeltaS^(@))/(R )` |
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