For reactions `P to Q` and `X to Y` Arrhenius constants are `10^(6) and 10^(8)` respectively. If `E_(P to Q) = 1500 cal//"mole" and E_(X to Y) = 2000 cal//"mole"`, then find the temperature at which their rate constant are same.
(Given : `R = 2 cal//"mole"//K)`

Options

A. 500 K
B. `250 xx 4.606 K`
C. `250/4.606 K`
D. `4.606/250 K`

Answer

To find the temperature at which the rate constants for the reactions \( P \to Q \) and \( X \to Y \) are the same, we can use the Arrhenius equation:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:
- \( k \) is the rate constant,
- \( A \) is the Arrhenius constant,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

### Step 1: Write the Arrhenius equations for both reactions

For the reaction \( P \to Q \):
\[ k = A_{PQ} e^{-\frac{E_{PQ}}{RT}} = 10^6 e^{-\frac{1500}{2T}} \]

For the reaction \( X \to Y \):
\[ k = A_{XY} e^{-\frac{E_{XY}}{RT}} = 10^8 e^{-\frac{2000}{2T}} \]

### Step 2: Set the rate constants equal to each other

Since we want to find the temperature at which the rate constants are the same:
\[ 10^6 e^{-\frac{1500}{2T}} = 10^8 e^{-\frac{2000}{2T}} \]

### Step 3: Simplify the equation

Dividing both sides by \( 10^6 \):
\[ e^{-\frac{1500}{2T}} = 10^2 e^{-\frac{2000}{2T}} \]

### Step 4: Rearranging the equation

Rearranging gives:
\[ e^{-\frac{1500}{2T}} = 100 e^{-\frac{2000}{2T}} \]

### Step 5: Take the natural logarithm of both sides

Taking the natural logarithm:
\[ -\frac{1500}{2T} = \ln(100) - \frac{2000}{2T} \]

### Step 6: Solve for \( T \)

Rearranging gives:
\[ -\frac{1500}{2T} + \frac{2000}{2T} = \ln(100) \]

This simplifies to:
\[ \frac{500}{2T} = \ln(100) \]
\[ \frac{250}{T} = \ln(100) \]

Using \( \ln(100) = 2 \ln(10) \) and \( \ln(10) \approx 2.303 \):
\[ \ln(100) \approx 4.606 \]

Thus,
\[ \frac{250}{T} = 4.606 \]

Now, solving for \( T \):
\[ T = \frac{250}{4.606} \approx 54.3 \text{ K} \]

### Final Answer
The temperature at which the rate constants are the same is approximately \( 54.3 \text{ K} \).

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