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 :

Options

A. 50
B. -50
C. 10
D. none of these

Answer

B