`A to A, DeltaH=+Ve`, Graph between `log_(10)A and 1/T` is straight line of slpe `1/(4.606)` find `DeltaH` in cel

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the value of ΔH in calories.

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
We start with the thermodynamic relationship:
\[ \Delta G = \Delta H - T \Delta S \]
And also, we know that:
\[ \Delta G = -2.303 R T \log K \]
Where \( K \) is the equilibrium constant.

### Step 2: Set the equations equal
By equating the two expressions for ΔG, we have:
\[ -2.303 R T \log K = \Delta H - T \Delta S \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \Delta H = T \Delta S - 2.303 R T \log K \]
Dividing through by T:
\[ \frac{\Delta H}{T} = \Delta S - 2.303 R \log K \]

### Step 4: Identify the slope of the graph
From the problem, we know that the graph of \( \log K \) versus \( \frac{1}{T} \) is a straight line. The slope \( m \) of this line can be expressed as:
\[ m = -\frac{\Delta H}{2.303 R} \]

### Step 5: Substitute the given slope
We are given that the slope \( m = \frac{1}{4.606} \). Thus, we can write:
\[ -\frac{\Delta H}{2.303 R} = \frac{1}{4.606} \]

### Step 6: Solve for ΔH
Rearranging gives:
\[ \Delta H = -2.303 R \cdot \frac{1}{4.606} \]

### Step 7: Substitute the value of R
The value of the gas constant \( R \) in calories is approximately \( 2 \, \text{cal/mol·K} \). Substituting this value in:
\[ \Delta H = -2.303 \cdot 2 \cdot \frac{1}{4.606} \]

### Step 8: Calculate ΔH
Calculating the numerical value:
\[ \Delta H = -\frac{2.303 \cdot 2}{4.606} \]
\[ \Delta H \approx -1.000 \, \text{cal/mol} \]

Since ΔH is given as positive in the problem, we take the absolute value:
\[ \Delta H \approx 1.000 \, \text{cal/mol} \]

### Final Answer
Thus, the value of ΔH in calories is:
\[ \Delta H \approx 1 \, \text{cal/mol} \]