If standard heat of dissociation of PCl...

If standard heat of dissociation of `PCl_(5)` is 230 cal than slope of the graph of `log k ` vs` (1)/(T) ` is `:`

a. `+50`

b. `-50`

c. `10`

d. None

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Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction DeltaG^(@)=-2.30RTlogk DeltaG^(@): Standing free energy change DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) …(ii) DeltaH^(@) : Standard heat of the reaction gt From eqns.(i) and(ii) -2RTlogk=DeltaH^(@)=TDeltaS^(@) DeltaS^(@) : standard entropy change implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R) Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope =(-DeltaH^(@))/(2.3R) amd y intercept =(DeltaS^(@))/(2.3R) If at temperature T_(1) equilibrium constant be k_(1) and at temperature T_(2) equilibrium constant be k_(2) then : implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R) ..(iv) implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R) ...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction If statndard heat of dissociation of PCl_(5) is 230 cal then slope of the graph of log vs (1)/(T) is :

The intercepts and slope of graph of log K_(p)^(@) "Vs" (1)/(T) are :

In a plot of log k vs 1/T, the slope is

In a plot of log.k vs 1/T, the slope'is :

For the Arrhenius equation, the slope for the plot ln k vs 1/T is

Which of the plots of ln K vs (1/T) is/are correct?

PCl_(5) on heating dissociates to PCl _(3) and Cl _(2) because :

The plot og log k vs 1//T helps to calculate

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