The variation of lnKc vs. `(l)/(T)` gives straight line having an angle of `45^(@)` . The value of heat of reaction in Cal is _____

Answer

To solve the problem, we will use the relationship between the equilibrium constant (Kc) and the temperature (T) as described by the van 't Hoff equation. The equation can be expressed in a linear form which allows us to determine the heat of reaction (ΔH) from the slope of the line.

### Step-by-Step Solution:

1. **Understanding the Equation**:
The van 't Hoff equation relates the change in the equilibrium constant (Kc) with temperature:
\[ \ln K_c = -\frac{\Delta H}{R} \cdot \frac{1}{T} + \Delta S \]
Here, ΔH is the heat of reaction, R is the universal gas constant, and ΔS is the change in entropy.

2. **Identifying the Linear Form**:
The equation can be rearranged to fit the linear equation format \(y = mx + b\):
\[ y = \ln K_c, \quad x = \frac{1}{T}, \quad m = -\frac{\Delta H}{R}, \quad b = \Delta S \]
This indicates that if we plot \(\ln K_c\) against \(\frac{1}{T}\), we will get a straight line.

3. **Analyzing the Slope**:
The problem states that the angle of the line is \(45^\circ\). The slope (m) of a line at \(45^\circ\) is given by:
\[ \tan(45^\circ) = 1 \]
Therefore, we have:
\[ m = -\frac{\Delta H}{R} = -1 \]

4. **Solving for ΔH**:
Rearranging the slope equation gives:
\[ -\Delta H = R \]
Thus,
\[ \Delta H = -R \]

5. **Substituting the Value of R**:
The value of the universal gas constant \(R\) in calories is approximately \(2\) cal/(mol·K). Therefore:
\[ \Delta H = -2 \text{ cal} \]

6. **Final Answer**:
The heat of reaction (ΔH) is:
\[ \Delta H = -2 \text{ cal} \]

### Final Answer:
The value of heat of reaction in Cal is **-2 Cal**.

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