1.

For the system at equlibrium, which of the following are correct

Answer»

`LOGK=(1)/(2.303R)(DeltaS-(DeltaH)/(T))`
On increasing the temperature of an endothemic reaction, the equlibrium shifts in forwared DIRECTION because Q decreases
On increasing the temperture of an endothermic reaction, the equilibrium shifts in forward direction because K increase
On increasing the temperature of an endothemic reaction, the conncentration in moles per LITRE of the reactants increase

Solution :`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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