Collision Theory & Reaction Rates│Please Help!

Question

Why does increasing the concentrations of the two aqueous solutions of a double replacement reaction increase the rate of reaction?

1. Higher concentrations will increase the temperature of the system, thus increasing the rate of reaction
2. Higher concentration means more moles of reactants in the same volume, which leads to more collisions
3. Increasing the concentrations will NOT impact the rate of reaction for a double replacement.
4. Higher concentrations will increase the activation energy of the system, thus increasing the rate of reaction

Answer 1

The answer is C:

1. Higher concentration means more moles of reactants in the same volume, which leads to more collisions

Consider the simple case of mixing equal volumes of 2 reactant solutions.

Experiment 1: Before mixing, the concentration of reactant A is C₁(A) = n₁(A) / V₁(A).

After mixing (in the instant before reaction), the concentration of A is

C_mx1(A) = n₁(A) / ( V₁(A)+V₁(B) )

but V₁(A) = V₁(B)

so C_mx(A) = n₁(A) / (2*V(A) ) = (1/2) * C₁(A)

Essentially, the moles of a don't change (YET) but mixing halved the concentration of the A reactant.

Experiment 2: Now double the concentration of reactants so that C₂(A) = 2*n₁(A) / V₁(A)..

After mixing, we again double the volume of reactant A, but now

C_mx2(A) = 2*n₁(A) / ( V₁(A)+V₁(B) ) = 2*n₁(A) / (2*V₁(A) ) = (2/2) * n₁(A) / V₁(A) = C₁(A)

Now the concentration of A after mixing is equal to the original concentration of A in Exp1 before mixing.

The mixed concentration of A in Exp2 is twice that if the mixed concentration of A in Exp1,

Doubling the initial concentration cancels the dilution from mixing 2 equal volumes; and similarly for reactant B.

Therefore the higher mixed concentrations of A &B result in higher rates.