in a reaction carried out at 500k ,0.001%of the total number of collisions are effective .The energy of activation of the reaction is approxiamately 1.15.8 kcal/mol 2.11.5 kcal/mol 3.12.8 kcal/mol 4.zero.

To determine the energy of activation for the reaction you mentioned, we can use the Arrhenius equation, which relates the rate constant of a reaction to the temperature and activation energy. The equation is expressed as:

k = A * e^(-Ea/RT)

Where:

• k = rate constant
• A = pre-exponential factor (frequency factor)
• e = base of the natural logarithm
• Ea = activation energy (in joules per mole)
• R = universal gas constant (8.314 J/mol·K)
• T = temperature in Kelvin

In your case, the reaction occurs at 500 K, and only 0.001% of the total collisions are effective. This means that the fraction of effective collisions can be represented as:

f = 0.001% = 0.00001

From the Arrhenius equation, we can derive that the fraction of effective collisions is related to the activation energy and temperature. The fraction of effective collisions can be approximated as:

f ≈ e^(-Ea/RT)

Taking the natural logarithm of both sides gives us:

ln(f) = -Ea/RT

Now, substituting the values we have:

ln(0.00001) = -Ea/(8.314 J/mol·K * 500 K)

Calculating the left side:

ln(0.00001) ≈ -11.5129

Now we can rearrange the equation to solve for Ea:

Ea = -ln(0.00001) * (8.314 J/mol·K * 500 K)

Substituting the values:

Ea ≈ 11.5129 * 8.314 * 500

Calculating this gives:

Ea ≈ 47857.5 J/mol

To convert this to kcal/mol, we divide by 4184 (since 1 kcal = 4184 J):

47857.5 J/mol ÷ 4184 J/kcal ≈ 11.43 kcal/mol

Rounding this value gives us approximately 11.5 kcal/mol. Therefore, the closest answer from your options is:

2. 11.5 kcal/mol

This calculation illustrates how the activation energy can be derived from the fraction of effective collisions and the temperature of the reaction. It highlights the importance of understanding the relationship between molecular collisions and reaction kinetics in chemistry.