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 for a particular reversible reaction
`K_(C)=57` abd `355^(@)C` and `K_(C)=69` at `450^(@)C` then

Options

A. `DeltaHlt0`
B. `DeltaHgt0`
C. `DeltaH=0`
D. `DeltaH` whose sign can't be determined

Answer

B

Explanation

Using equation, `"log"(K_(2))/(K_(1))=(DeltaH^(@))/(2.30R)((1)/(T_(1))-(1)/(T_(2)))`