A particular chemical reaction occurs at room temperature (293 K) at half the rate that it does at 300 K. What is the activation energy for this reaction?

To solve this problem, we can use the Arrhenius equation which describes the temperature dependence of reaction rates:

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)),
• T is the temperature in Kelvin.

Given that the reaction rate at 293 K is half of that at 300 K, we can set up the following equation:

k1/k2 = e^((Ea/R)(1/T1 - 1/T2))

where:

• k1 is the rate constant at T1 (293 K),
• k2 is the rate constant at T2 (300 K).

Since k1 is half of k2, we can substitute k1/k2 with 0.5:

0.5 = e^((Ea/R)(1/293 - 1/300))

To solve for Ea, we first take the natural logarithm of both sides:

ln(0.5) = (Ea/R)(1/293 - 1/300)

Then we can solve for Ea:

Ea = R * ln(0.5) / (1/293 - 1/300)

Substituting R with 8.314 J/(mol·K), we get:

Ea = 8.314 * ln(0.5) / (1/293 - 1/300)

Calculating the above expression will give us the activation energy in Joules per mole.