1. Use collision theory and Arrhenius form

For a bimolecular reaction, the experimentally observed rate constant can be written as

$k = P Z e^{-E_a/RT}$

where:

• $Z$ = collision frequency,
• $P$ = steric factor,
• $E_a$ = activation energy.

In Arrhenius form,

$k = A e^{-E_a/RT}$

So,

$A = PZ$

Thus the steric factor affects the frequency factor $A$, not the activation energy $E_a$.

2. Given $P = 4.5$

Since

$A = PZ = 4.5Z$

we have

$A > Z$

That means the experimentally determined frequency factor is greater than the simple collision-frequency prediction.

3. Check each option

Option A:

The activation energy of the reaction is unaffected by the value of the steric factor

True, because $P$ appears in the pre-exponential factor:

$A = PZ$

and does not change $E_a$.

✅ A is correct

Option B:

Experimentally determined value of frequency factor is higher than that predicted by arrhenius equation

The intended comparison is between the experimentally observed pre-exponential factor and the basic collision prediction $Z$.

Since

$A = PZ = 4.5Z,$

experimental $A$ is higher than the collision-based prediction.

✅ B is correct

Option C:

Since $P = 4.5,$ the reaction will not proceed unless an effective catalyst is used

This is false. A steric factor greater than $1$ does not imply the reaction cannot proceed. It only means the effective pre-exponential factor is larger than the naive collision estimate.

❌ C is incorrect

Option D:

The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

This is opposite to what we found. Since $P = 4.5 > 1$,

$A = 4.5Z > Z$

So the experimentally determined frequency factor is higher, not lower.

❌ D is incorrect

4. Final answer

The correct options are:

$\boxed{A,\ B}$

5. Comparison with stored correct answer

Stored correct answer: B, A

This matches our derived answer exactly.