For a particular chemical reaction, the rate constant at 30.0 °C is 1.38 X 10-4 L mol-1 s-1, while the value at 49.0 °C is 1.21 X 10-3 L mol-1 s-1 . What is the activation energy for this reaction?
a. 92.8 kJ
b. 200 kJ
c. 40.4 kJ
d. 343 kJ
e. 56.4 kJ
ln 1.38x 10 -4 /
1.21 x 10 - 3 = Ea (1/303) - (1/322) =1.9473
8.70= 8.314 /
1.95 x 10 ^ -4
thats all I know how to do help please!
To find the activation energy for this reaction, you can use the Arrhenius equation:
k = Ae^(-Ea/RT),
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
Given that the rate constant at 30.0 °C is 1.38 x 10^-4 L mol^-1 s^-1 (298 K) and the rate constant at 49.0 °C is 1.21 x 10^-3 L mol^-1 s^-1 (322 K), we can set up the equation as follows:
1.38 x 10^-4 = A * e^(-Ea/(8.314 * 298))
1.21 x 10^-3 = A * e^(-Ea/(8.314 * 322))
To eliminate A, we can divide the two equations:
(1.38 x 10^-4) / (1.21 x 10^-3) = e^(-Ea/(8.314 * 298))/(e^(-Ea/(8.314 * 322)))
1.1405 = e^(-Ea/2470.972) / e^(-Ea/2662.948)
Next, we take the natural logarithm of both sides to obtain:
ln(1.1405) = -Ea/2470.972 + Ea/2662.948
Now, we can simplify the equation:
ln(1.1405) = Ea / (8.314) * (1/2662.948 - 1/2470.972)
ln(1.1405) = Ea / 8.314 * (0.0003786711 - 0.0004070957)
ln(1.1405) = Ea / 8.314 * (-0.0000284246)
ln(1.1405) = -0.0000284246 * Ea / 8.314
Now, we can solve for Ea by isolating it on one side of the equation:
ln(1.1405) * 8.314 = -0.0000284246 * Ea
Ea = (ln(1.1405) * 8.314) / (-0.0000284246)
Ea ≈ -92.8 kJ
Therefore, the activation energy for this reaction is approximately -92.8 kJ.
You have reversed k1 and k2 and T1 and T2 and you don't have R anywhere in the equation.
k1 = 1.38E-4, 30C or 303K
k2 = 1.21E-3, 49C or 322K
ln(k2/k1) = (1/303)-(1/322)*Ea/R
ln(1.21-3/1.38E-4) = (.00330 - 0.00311)*Ea/8314
ln 8.768 = 0.000195Ea/8.314
2.171 = 0.0000234*Ea
Ea = ?J