The rate constant of a chemical equation has increased from 0.100s^-1 to 2.70s^-1 upon raising the temperature from 25C to 45C. Find the activation energy.
So far I done 3.30=(Ea/8.314)(-2.11x10^-4)
To determine the activation energy (Ea), we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the activation energy:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant,
- A is the pre-exponential factor (a constant),
- Ea is the activation energy,
- R is the gas constant (8.314 J/(mol*K)), and
- T is the temperature in Kelvin.
Given:
k1 = 0.100 s^(-1) (rate constant at 25°C = 298.15 K)
k2 = 2.70 s^(-1) (rate constant at 45°C = 318.15 K)
We can rewrite the Arrhenius equation for the two temperatures:
k1 = A * e^(-Ea/(8.314 * 298.15))
k2 = A * e^(-Ea/(8.314 * 318.15))
To eliminate the pre-exponential factor (A), we can divide the two equations:
k2/k1 = e^(-Ea/(8.314 * 318.15)) / e^(-Ea/(8.314 * 298.15))
Simplifying further:
k2/k1 = e^(-Ea/8.314) * e^(Ea/(8.314 * 298.15))
Since the same activation energy (Ea) is common to both terms on the right side, we can simplify by canceling them out:
k2/k1 = e^(Ea/(8.314 * 318.15 - 8.314 * 298.15))
k2/k1 = e^(Ea/8.314 * (318.15 - 298.15))
k2/k1 = e^(Ea/8.314 * 20)
Now we can substitute the given values:
2.70/0.100 = e^(Ea/(8.314 * 20))
To solve for Ea, take the natural logarithm (ln) of both sides:
ln(2.70/0.100) = Ea/(8.314*20)
Now we can rearrange and solve for Ea:
Ea = ln(2.70/0.100) * (8.314*20)
Calculating:
Ea ≈ 46.08 J/mol
So, the activation energy is approximately 46.08 J/mol.