An experiment was performed twice at different temperatures. If the value of the rate constant for the reaction doubles when the temperature is increased from 17°c to 27°c, determine the activation energy for the reaction?
To determine the activation energy for the reaction, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/mol*K)
T = temperature in Kelvin
Given that the rate constant doubles when the temperature increases from 17°C to 27°C, we can set up the following relationship:
k2/k1 = 2,
where k2 is the rate constant at 27°C and k1 is the rate constant at 17°C. Therefore, we have:
( A * e^(-Ea/(R * 300 K)) ) / ( A * e^(-Ea/(R * 290 K) ) = 2
Simplifying the above equation, we get:
e^(Ea/(8.314 * 300)) / e^(Ea/(8.314 * 290)) = 2
e^[(Ea/8.314) * (1/300 - 1/290)] = 2
e^[(Ea/8.314) * (290 - 300)/(290*300)] = 2
e^(Ea/8.314 * (-10)/(8700)) = 2
e^(-Ea/870) = 2
Taking the natural logarithm of both sides:
ln(e^(-Ea/870)) = ln(2)
-Ea/870 = ln(2)
Ea = -870 * ln(2)
Ea ≈ 6025 J/mol
Therefore, the activation energy for the reaction is approximately 6025 J/mol.