The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?
Given that the initial rate constant is 0.0160 s-1 at an initial temperature of 25 degrees C , what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?
What's your trouble with this?
b.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
ln(k2/0.0160) = (32900/0.08206)(1/298/15 - 1/403.15)
Solve for k2.
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To determine the temperature at which the reaction would go twice as fast, we can use the Arrhenius equation:
k2/k1 = e^[-(Ea/R) * (1/T2 - 1/T1)]
Where:
k1 and k2 are the rate constants at temperatures T1 and T2 respectively,
Ea is the activation energy,
R is the gas constant (8.314 J/mol*K),
T1 and T2 are the temperatures in Kelvin.
Let's begin with the first part of the question:
1. Calculate T1 in Kelvin:
T1 = 25 degrees Celsius + 273.15 = 298.15 K
2. Set up the equation using the given information:
2 = e^[-(32.9 kJ/mol / (8.314 J/mol*K)) * (1/T2 - 1/298.15 K)]
3. Solve for T2:
To do this, we need to isolate the exponential term on one side of the equation and then take the natural logarithm (ln) of both sides.
ln(2) = -(32.9 kJ/mol / (8.314 J/mol*K)) * (1/T2 - 1/298.15 K)
4. Rearrange the equation and solve for 1/T2:
-(32.9 kJ/mol / (8.314 J/mol*K)) * (1/T2 - 1/298.15 K) = ln(2)
-(32.9 kJ/mol / (8.314 J/mol*K)) * 1/T2 + (32.9 kJ/mol / (8.314 J/mol*K)) * 1/298.15 K = ln(2)
-(32.9 kJ/mol / (8.314 J/mol*K)) * 1/T2 = ln(2) - (32.9 kJ/mol / (8.314 J/mol*K)) * 1/298.15 K
1/T2 = -((8.314 J/mol*K) / 32.9 kJ/mol) * (ln(2) - (32.9 kJ/mol / (8.314 J/mol*K)) * 1/298.15 K)
T2 = 1 / [((8.314 J/mol*K) / 32.9 kJ/mol) * (ln(2) - (32.9 kJ/mol / (8.314 J/mol*K)) * 1/298.15 K)]
Now we can calculate T2 using this equation.
For the second part of the question:
1. Calculate T1 and T2 in Kelvin:
T1 = 25 degrees Celsius + 273.15 = 298.15 K
T2 = 130 degrees Celsius + 273.15 = 403.15 K
2. Set up the equation using the same form of the Arrhenius equation:
k2/k1 = e^[-(Ea/R) * (1/T2 - 1/T1)]
3. Solve for k2:
k2 = k1 * e^[-(Ea/R) * (1/T2 - 1/T1)]
Substitute the given values of k1 and T1 into the equation to calculate k2.