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

Explanation

To find the slope of the graph of log K versus 1/T, we can follow these steps:

### Step 1: Understand the relationship
We start with the equation derived from thermodynamics:
\[ \log K = -\frac{\Delta H^\circ}{2.303R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{2.303R} \]
This equation is in the form of \( y = mx + c \), where:
- \( y = \log K \)
- \( m = -\frac{\Delta H^\circ}{2.303R} \) (the slope)
- \( x = \frac{1}{T} \)
- \( c = \frac{\Delta S^\circ}{2.303R} \) (the y-intercept)

### Step 2: Identify given values
From the problem, we know that the standard heat of dissociation of \( PCl_5 \) is given as:
\[ \Delta H^\circ = 230 \text{ cal} \]
We also need to convert calories to joules since the gas constant \( R \) is usually in joules. The conversion factor is:
\[ 1 \text{ cal} = 4.184 \text{ J} \]
Thus,
\[ \Delta H^\circ = 230 \text{ cal} \times 4.184 \text{ J/cal} = 962.32 \text{ J} \]

### Step 3: Calculate the slope
Now we can substitute the values into the slope formula:
\[ m = -\frac{\Delta H^\circ}{2.303R} \]
Using \( R = 8.314 \text{ J/(mol K)} \):
\[ m = -\frac{962.32 \text{ J}}{2.303 \times 8.314 \text{ J/(mol K)}} \]

Calculating the denominator:
\[ 2.303 \times 8.314 \approx 19.1 \text{ J/(mol K)} \]

Now substituting this back into the slope equation:
\[ m = -\frac{962.32}{19.1} \approx -50.4 \]

### Step 4: Final result
Thus, the slope of the graph of log K versus \( \frac{1}{T} \) is approximately:
\[ m \approx -50 \]

### Conclusion
The slope of the graph of log K versus \( \frac{1}{T} \) is -50.

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