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The activation energy for the reaction -...

The activation energy for the reaction -
`H_(2)O_(2)toH_(2)O+(1)/(2)O_(2)`
is `18 K cal//mol` at 300 K. calculate the fraction of molecules of reactonts having energy equal to or greater than activation energy ?
Anti log `(-13.02)=9.36xx10^(-14)`

A

`9.36xx10^(-14)`

B

`1.2xx10^(-12)`

C

`4.2xx10^(-16)`

D

`5.2xx10^(-15)`

Text Solution

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The correct Answer is:
To solve the problem of calculating the fraction of molecules of reactants having energy equal to or greater than the activation energy for the reaction \( H_2O_2 \to H_2O + \frac{1}{2} O_2 \) with an activation energy \( E_a = 18 \, \text{kcal/mol} \) at \( T = 300 \, K \), we will use the Arrhenius equation. ### Step-by-Step Solution: 1. **Understand the Formula**: The fraction of molecules \( F \) having energy equal to or greater than the activation energy is given by: \[ F = e^{-\frac{E_a}{RT}} \] where: - \( E_a \) is the activation energy, - \( R \) is the universal gas constant, and - \( T \) is the temperature in Kelvin. 2. **Convert Activation Energy**: Since the activation energy is given in kilocalories, we need to convert it to calories: \[ E_a = 18 \, \text{kcal/mol} \times 1000 \, \text{cal/kcal} = 18000 \, \text{cal/mol} \] 3. **Identify the Gas Constant**: The value of \( R \) in appropriate units (calories) is: \[ R = 2 \, \text{cal/(mol K)} \] 4. **Substitute Values into the Formula**: Now we can substitute the values into the fraction formula: \[ F = e^{-\frac{18000}{2 \times 300}} \] 5. **Calculate the Exponent**: First, calculate the denominator: \[ 2 \times 300 = 600 \] Now calculate the exponent: \[ -\frac{18000}{600} = -30 \] 6. **Calculate the Fraction**: Now we can compute \( F \): \[ F = e^{-30} \] 7. **Use the Anti-logarithm**: To express this in terms of a numerical value, we can use the approximation: \[ e^{-30} \approx 10^{-30} \] (This approximation is valid since \( e^{-x} \) can be estimated as \( 10^{-x \cdot \log_{10} e} \), and \( \log_{10} e \approx 0.434 \)). 8. **Final Result**: Thus, the fraction of molecules of reactants having energy equal to or greater than the activation energy is: \[ F \approx 10^{-30} \]

To solve the problem of calculating the fraction of molecules of reactants having energy equal to or greater than the activation energy for the reaction \( H_2O_2 \to H_2O + \frac{1}{2} O_2 \) with an activation energy \( E_a = 18 \, \text{kcal/mol} \) at \( T = 300 \, K \), we will use the Arrhenius equation. ### Step-by-Step Solution: 1. **Understand the Formula**: The fraction of molecules \( F \) having energy equal to or greater than the activation energy is given by: \[ F = e^{-\frac{E_a}{RT}} ...
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