Question

Variation of \(\log_{10} K\) with \(\dfrac{1}{T}\) is shown by the following graph in which straight line is at \(45^\circ\). Hence \(\Delta H^\circ\) is:

• A. \(+4.606\,\text{cal}\)
• B. \(-4.606\,\text{cal}\)
• C. \(2\,\text{cal}\)
• D. \(-2\,\text{cal}\)

Hint:

In a \(\log K\) vs \(\dfrac{1}{T}\) plot:
Positive slope \(\Rightarrow \Delta H^\circ<0\) (exothermic reaction)
Negative slope \(\Rightarrow \Delta H^\circ>0\) (endothermic reaction)

Answer 1

The Correct Option is B

Step 1: The van’t Hoff equation relating equilibrium constant and temperature is: \[ \log_{10} K = -\frac{\Delta H^\circ}{2.303\,R}\left(\frac{1}{T}\right) + \text{constant} \]
Step 2: Hence, the slope of the graph between \(\log_{10} K\) and \(\dfrac{1}{T}\) is: \[ \text{slope} = -\frac{\Delta H^\circ}{2.303\,R} \]
Step 3: The straight line is given to be at \(45^\circ\), so: \[ \text{slope} = \tan 45^\circ = 1 \]
Step 4: Substituting: \[ 1 = -\frac{\Delta H^\circ}{2.303\,R} \] \[ \Delta H^\circ = -2.303\,R \]
Step 5: Using \(R = 1.987\,\text{cal mol}^{-1}\text{K}^{-1}\): \[ \Delta H^\circ = -2.303 \times 1.987 \approx -4.606\,\text{cal} \]