2 NO + O2 --> 2 NO2

The reaction rate is increased five­fold when the temperature is 
increased from 1400 K to 1500 K. What is the activation energy for the reaction? 

I have no idea how to do this and would greatly appreciate some urgent help!

To determine the activation energy for the reaction, we can use the Arrhenius equation:

k = Ae^(-Ea/RT)

Where:
- k is the rate constant for the reaction
- A is the pre-exponential factor (related to the frequency of collision and orientation of reactant molecules)
- Ea is the activation energy
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin

In this case, we want to compare the rate at two different temperatures, 1400 K and 1500 K. Let's assume that k1 and k2 are the rate constants at those temperatures, respectively. We can express this as:

k1 = A * e^(-Ea/RT1)
k2 = A * e^(-Ea/RT2)

Since we know that the reaction rate is increased five-fold when the temperature increases from 1400 K to 1500 K, we can write this as:

k2 = 5 * k1

Dividing the second equation by the first equation, we get:

k2 / k1 = (A * e^(-Ea/RT2)) / (A * e^(-Ea/RT1))
5 = e^(-Ea/RT2) / e^(-Ea/RT1)
5 = e^(-Ea/RT2 + Ea/RT1)

Taking the natural logarithm (ln) of both sides:

ln(5) = -Ea/RT2 + Ea/RT1

Rearranging the equation:

Ea/RT1 - Ea/RT2 = ln(5)

Factoring out the activation energy (Ea):

Ea * (1/RT1 - 1/RT2) = ln(5)

Now, plug in the given temperatures, R, and solve for Ea:

Ea = (ln(5)) / ((1/RT1) - (1/RT2))

Substituting the values:
- Temperature (T1) = 1400 K
- Temperature (T2) = 1500 K
- Ideal gas constant (R) = 8.314 J/(mol·K)

Ea = (ln(5)) / ((1/(8.314 J/(mol·K) * 1400 K)) - (1/(8.314 J/(mol·K) * 1500 K)))

By calculating this expression, we can find the activation energy for the reaction.