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

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Relationship**: We start with the relationship between Gibbs free energy change (ΔG), enthalpy change (ΔH), and entropy change (ΔS):
\[ \Delta G = \Delta H - T\Delta S \]
and the relationship between ΔG and the equilibrium constant (K):
\[ \Delta G = -RT \ln K \]

2. **Equating the Two Expressions**: By equating the two expressions for ΔG, we have:
\[ -RT \ln K = \Delta H - T\Delta S \]
Rearranging gives:
\[ \ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R} \]

3. **Considering Two Different Temperatures**: Let’s denote the equilibrium constants at two different temperatures \(T_1\) and \(T_2\) as \(K_1\) and \(K_2\):
\[ \ln K_1 = -\frac{\Delta H}{RT_1} + \frac{\Delta S}{R} \]
\[ \ln K_2 = -\frac{\Delta H}{RT_2} + \frac{\Delta S}{R} \]

4. **Subtracting the Two Equations**: We subtract the first equation from the second:
\[ \ln K_2 - \ln K_1 = -\frac{\Delta H}{RT_2} + \frac{\Delta H}{RT_1} \]
This can be rewritten using properties of logarithms:
\[ \ln \left(\frac{K_2}{K_1}\right) = -\Delta H \left(\frac{1}{RT_2} - \frac{1}{RT_1}\right) \]

5. **Calculating the Values**: Given \(K_1 = 57\) at \(T_1 = 355^\circ C\) and \(K_2 = 69\) at \(T_2 = 450^\circ C\), we first convert temperatures to Kelvin:
- \(T_1 = 355 + 273 = 628 \, K\)
- \(T_2 = 450 + 273 = 723 \, K\)

6. **Plugging in the Values**: Substitute the values into the equation:
\[ \ln \left(\frac{69}{57}\right) = -\Delta H \left(\frac{1}{8.314 \times 723} - \frac{1}{8.314 \times 628}\right) \]

7. **Calculating the Left Side**: Calculate \( \ln \left(\frac{69}{57}\right) \):
\[ \ln \left(\frac{69}{57}\right) \approx 0.19105 \]

8. **Calculating the Right Side**: Calculate the difference:
\[ \frac{1}{723} - \frac{1}{628} \approx -0.000131 \]
Now, substituting back:
\[ 0.19105 = -\Delta H \left(-0.000131\right) \cdot \frac{1}{8.314} \]

9. **Solving for ΔH**: Rearranging gives:
\[ \Delta H = \frac{0.19105 \cdot 8.314}{0.000131} \]
After calculation, we find:
\[ \Delta H \approx 3.10 \times 10^3 \, J/mol \]

10. **Conclusion**: Since ΔH is positive, it indicates that the reaction is endothermic. Thus, the equilibrium constant increases with an increase in temperature.