Answer:
K2 = 61.2 M^-1.S^-1
Explanation:
We complete the question fully:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
Answer is as follows:
The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:
According to Arrhenius equation, the relationship between temperature and activation energy is as follows:
k = Ae^-(Ea/RT)
where, k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant
T = temperature in kelvin
From the equation, the following was derived for a double temperature problem:
ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)
We list out the parameters as follows:
T1= (244 + 273.15) K = 517.15 K
T2= (324+ 273.15) K =597.15 K
K1 = 6.7 , K2 = ?
R = 8.314 J/mol K
Ea = 71.0 kJ/mol = 71000 J/mol
Putting the given values into the above formula as follows:
ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)
lnk2 - 1.902 = 8539.8 * 0.000259
lnK2 = 1.902 + 2.21
lnK2 = 4.114
K2 = e^(4.114)
K2 = 61.2
Hence, K2 = 61.2 (M.S)^-1
