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
For exothermic reaction if `DeltaS^(@)lt0` then the sketch of log k vs `(1)/(T)` may be

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Answer

To analyze the effect of temperature on the equilibrium constant using thermodynamics, we can follow these steps:

### Step 1: Understand the Relationship Between Gibbs Free Energy and Equilibrium Constant
The Gibbs free energy change (ΔG°) for a reaction is related to the equilibrium constant (K) by the equation:
\[ \Delta G^(@) = -2.303RT \log K \]
This equation shows that the Gibbs free energy change is directly related to the logarithm of the equilibrium constant.

### Step 2: Relate Gibbs Free Energy to Enthalpy and Entropy
The Gibbs free energy change can also be expressed in terms of enthalpy (ΔH°) and entropy (ΔS°) changes:
\[ \Delta G^(@) = \Delta H^(@) - T \Delta S^(@) \]
This equation indicates that the change in Gibbs free energy is influenced by both the enthalpy change and the temperature multiplied by the entropy change.

### Step 3: Combine the Two Equations
By equating the two expressions for ΔG°:
\[ -2.303RT \log K = \Delta H^(@) - T \Delta S^(@) \]
Rearranging gives:
\[ \log K = -\frac{\Delta H^(@)}{2.303RT} + \frac{\Delta S^(@)}{2.303R} \]
This equation shows how the equilibrium constant K varies with temperature (T).

### Step 4: Analyze the Slope and Intercept
If we plot log K versus \( \frac{1}{T} \), we can identify the slope and intercept:
- The slope (m) is given by:
\[ m = -\frac{\Delta H^(@)}{2.303R} \]
- The y-intercept (c) is:
\[ c = \frac{\Delta S^(@)}{2.303R} \]
For an exothermic reaction, ΔH° is negative, which means the slope will be positive.

### Step 5: Consider the Sign of ΔS°
If ΔS° is negative, the intercept will also be negative. Thus, we expect a graph where:
- The slope is positive (due to negative ΔH°).
- The intercept is negative (due to negative ΔS°).

### Step 6: Conclusion
From the analysis, we conclude that for an exothermic reaction with ΔS° < 0, the plot of log K versus \( \frac{1}{T} \) will show a positive slope and a negative intercept.