InterviewSolution
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InterviewSolution
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For the system at equlibrium, which of the following are correctA. `logK=(1)/(2.303R)(DeltaS-(DeltaH)/(T))`B. On increasing the temperature of an endothemic reaction, the equlibrium shifts in forwared direction because Q decreasesC. On increasing the temperture of an endothermic reaction, the equilibrium shifts in forward direction because K increaseD. On increasing the temperature of an endothemic reaction, the conncentration in moles per litre of the reactants increase |
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Answer» Correct Answer - A::C `DeltaG=DeltaH-TDeltaS=-2.303RTlogK` `TDeltaS-DeltaH=2.303RTlogKimplies` `(TDeltaS-DeltaH)/(2.303RT)=logKimplies(1)/(1.303R)[DeltaS-(DeltaH)/(T)]=logK` Hence choice (a) is correct. On changing temperatue K changes and not Q. Also `log""(K_(2))/(K_(1))=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` For endothemic reactions on increasing T, K increase hte concentration of products. Thus choice (c) is correct while (b) and (d) are not correct. |
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