Answer :
To calculate the temperature at which the reaction rate will be ten times faster than the rate at [tex]\(0^\circ C\)[/tex] for a reaction with an activation energy of [tex]\(80 \, \text{kJ/mol}\)[/tex], we can use the Arrhenius equation. The steps involved in solving this problem are as follows:
1. Convert the Activation Energy to J/mol:
The given activation energy is [tex]\(80 \, \text{kJ/mol}\)[/tex]. Convert this value to [tex]\( \text{J/mol} \)[/tex] because the universal gas constant [tex]\( R \)[/tex] is typically expressed in [tex]\( \text{J/(mol·K)}\)[/tex].
[tex]\[ E_a = 80 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 80000 \, \text{J/mol} \][/tex]
2. Convert the Initial Temperature from Celsius to Kelvin:
The initial temperature is [tex]\(0^\circ C\)[/tex].
[tex]\[ T_1 = 0^\circ C + 273.15 = 273.15 \, \text{K} \][/tex]
3. Use the Arrhenius Equation:
The Arrhenius equation relates the rate constant [tex]\( k \)[/tex] to the temperature [tex]\( T \)[/tex]:
[tex]\[ k = A \cdot \exp\left(\frac{-E_a}{R \cdot T}\right) \][/tex]
Where [tex]\( A \)[/tex] is the pre-exponential factor, [tex]\( E_a \)[/tex] is the activation energy, [tex]\( R \)[/tex] is the universal gas constant [tex]\( (8.314 \, \text{J/(mol·K)}) \)[/tex], and [tex]\( T \)[/tex] is the temperature in Kelvin.
4. Express the Ratio of the Rates:
We know that at the new temperature [tex]\( T_2 \)[/tex], the rate will be ten times the rate at [tex]\( 0^\circ C \)[/tex]:
[tex]\[ \frac{k_2}{k_1} = 10 \][/tex]
By using the Arrhenius equation, we can write:
[tex]\[ \frac{k_2}{k_1} = \frac{A \cdot \exp\left(\frac{-E_a}{R \cdot T_2}\right)}{A \cdot \exp\left(\frac{-E_a}{R \cdot T_1}\right)} = \exp\left(\frac{-E_a}{R \cdot T_2}\right) / \exp\left(\frac{-E_a}{R \cdot T_1}\right) \][/tex]
5. Simplify the Exponential Terms:
[tex]\[ \frac{k_2}{k_1} = \exp\left(\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right) \][/tex]
Since [tex]\(\frac{k_2}{k_1} = 10\)[/tex]:
[tex]\[ 10 = \exp\left(\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right) \][/tex]
6. Take the Natural Logarithm of Both Sides:
[tex]\[ \ln(10) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \][/tex]
Given that [tex]\(\ln(10) \approx 2.302585\)[/tex]:
[tex]\[ 2.302585 = \frac{80000}{8.314} \left(\frac{1}{273.15} - \frac{1}{T_2}\right) \][/tex]
7. Solve for [tex]\( \frac{1}{T_2} \)[/tex]:
[tex]\[ 2.302585 = 9616.0508 \left(\frac{1}{273.15} - \frac{1}{T_2}\right) \][/tex]
[tex]\[ \frac{1}{T_2} = \frac{1}{273.15} - \frac{2.302585}{9616.0508} \][/tex]
Calculate [tex]\( \frac{1}{T_1} - \frac{\ln(10)}{E_a/R} \)[/tex]:
[tex]\[ \frac{1}{273.15} - \frac{2.302585}{9616.0508} = \approx 0.0034217 \][/tex]
8. Calculate [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = \frac{1}{0.0034217} = 292.2527 \, \text{K} \][/tex]
9. Convert Back to Celsius:
[tex]\[ T_2 = 292.2527 \, \text{K} - 273.15 \approx 19.10^\circ C \][/tex]
Therefore, the temperature at which the rate of the reaction will be ten times faster than that at [tex]\(0^\circ C\)[/tex] is approximately [tex]\(292.25 \, \text{K}\)[/tex] or [tex]\(19.10^\circ C\)[/tex].
1. Convert the Activation Energy to J/mol:
The given activation energy is [tex]\(80 \, \text{kJ/mol}\)[/tex]. Convert this value to [tex]\( \text{J/mol} \)[/tex] because the universal gas constant [tex]\( R \)[/tex] is typically expressed in [tex]\( \text{J/(mol·K)}\)[/tex].
[tex]\[ E_a = 80 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 80000 \, \text{J/mol} \][/tex]
2. Convert the Initial Temperature from Celsius to Kelvin:
The initial temperature is [tex]\(0^\circ C\)[/tex].
[tex]\[ T_1 = 0^\circ C + 273.15 = 273.15 \, \text{K} \][/tex]
3. Use the Arrhenius Equation:
The Arrhenius equation relates the rate constant [tex]\( k \)[/tex] to the temperature [tex]\( T \)[/tex]:
[tex]\[ k = A \cdot \exp\left(\frac{-E_a}{R \cdot T}\right) \][/tex]
Where [tex]\( A \)[/tex] is the pre-exponential factor, [tex]\( E_a \)[/tex] is the activation energy, [tex]\( R \)[/tex] is the universal gas constant [tex]\( (8.314 \, \text{J/(mol·K)}) \)[/tex], and [tex]\( T \)[/tex] is the temperature in Kelvin.
4. Express the Ratio of the Rates:
We know that at the new temperature [tex]\( T_2 \)[/tex], the rate will be ten times the rate at [tex]\( 0^\circ C \)[/tex]:
[tex]\[ \frac{k_2}{k_1} = 10 \][/tex]
By using the Arrhenius equation, we can write:
[tex]\[ \frac{k_2}{k_1} = \frac{A \cdot \exp\left(\frac{-E_a}{R \cdot T_2}\right)}{A \cdot \exp\left(\frac{-E_a}{R \cdot T_1}\right)} = \exp\left(\frac{-E_a}{R \cdot T_2}\right) / \exp\left(\frac{-E_a}{R \cdot T_1}\right) \][/tex]
5. Simplify the Exponential Terms:
[tex]\[ \frac{k_2}{k_1} = \exp\left(\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right) \][/tex]
Since [tex]\(\frac{k_2}{k_1} = 10\)[/tex]:
[tex]\[ 10 = \exp\left(\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right) \][/tex]
6. Take the Natural Logarithm of Both Sides:
[tex]\[ \ln(10) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \][/tex]
Given that [tex]\(\ln(10) \approx 2.302585\)[/tex]:
[tex]\[ 2.302585 = \frac{80000}{8.314} \left(\frac{1}{273.15} - \frac{1}{T_2}\right) \][/tex]
7. Solve for [tex]\( \frac{1}{T_2} \)[/tex]:
[tex]\[ 2.302585 = 9616.0508 \left(\frac{1}{273.15} - \frac{1}{T_2}\right) \][/tex]
[tex]\[ \frac{1}{T_2} = \frac{1}{273.15} - \frac{2.302585}{9616.0508} \][/tex]
Calculate [tex]\( \frac{1}{T_1} - \frac{\ln(10)}{E_a/R} \)[/tex]:
[tex]\[ \frac{1}{273.15} - \frac{2.302585}{9616.0508} = \approx 0.0034217 \][/tex]
8. Calculate [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = \frac{1}{0.0034217} = 292.2527 \, \text{K} \][/tex]
9. Convert Back to Celsius:
[tex]\[ T_2 = 292.2527 \, \text{K} - 273.15 \approx 19.10^\circ C \][/tex]
Therefore, the temperature at which the rate of the reaction will be ten times faster than that at [tex]\(0^\circ C\)[/tex] is approximately [tex]\(292.25 \, \text{K}\)[/tex] or [tex]\(19.10^\circ C\)[/tex].
