Calculate Activation Energy, Ea, for each of the reactions, above? ln(k)= -2.24x10^4(1/T)-1.8. The reactions are Step 1: 2H2 + N2 -> N2H4, Step 2: N2H4 + H2 -> 2NH3, Overall: 3H2 + N2 -> NH3. This is all the information i have. i Don't know how to solve for the Activation energy using just this?

To calculate the activation energy (Ea) using the given equation ln(k) = -2.24x10^4(1/T) -1.8, we can use the Arrhenius equation which relates the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea).

The Arrhenius equation is given by:
k = Ae^(-Ea/RT)

Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/mol·K)
T = absolute temperature (in Kelvin)

In the given equation, we have ln(k) = -2.24x10^4(1/T) - 1.8.
By comparing this with the Arrhenius equation, we can see that the slope of ln(k) versus 1/T corresponds to -Ea/R.

So, we can determine the activation energy by solving for -Ea/R. Let's calculate it step by step:

Step 1: Rearrange the equation:
ln(k) = -2.24x10^4(1/T) - 1.8
ln(k) + 1.8 = -2.24x10^4(1/T)

Step 2: Convert the natural logarithm to exponential form:
k = e^(-2.24x10^4(1/T) - 1.8)

Step 3: Take the exponent of both sides:
e^(ln(k) + 1.8) = e^(-2.24x10^4(1/T))
e^(ln(k)) * e^(1.8) = e^(-2.24x10^4(1/T))

Step 4: Simplify:
k * e^(1.8) = e^(-2.24x10^4(1/T))

Step 5: Cancel out the exponential function on one side:
k = e^(-2.24x10^4(1/T) - 1.8)/e^(1.8)

Step 6: Further simplification:
k = e^(-2.24x10^4(1/T) - 1.8 + 1.8)
k = e^(-2.24x10^4(1/T))

Step 7: Comparing with the Arrhenius equation:
We now have k = Ae^(-Ea/RT)

Comparing the two equations, we observe that the pre-exponential factor (A) is equal to 1, and the term -Ea/RT is equal to -2.24x10^4(1/T).

Step 8: Solve for -Ea/RT:
-2.24x10^4(1/T) = -Ea/RT

Step 9: Cross-multiply and solve for Ea:
-Ea/RT = -2.24x10^4

Ea = 2.24x10^4 * R

Now you can substitute the value of the gas constant (R = 8.314 J/mol·K) into the equation to calculate the activation energy (Ea).