The gas-phase reaction, A2 + B2 --> 2AB, proceeds by bimolecular collisions between A2 and B2 molecules. If the concentrations of both A2 and B2 are doubled, the reaction rate will change by a factor of.....?

Will you please explain instead of just giving me an answer. thanks

Hey! amarillo science on youtube does a great job at explaining it:

basically, the rate eq'n is: rate = k [A][B]

Note: we know that k will always stay the same since it's a constant

so, if the concentrations were, for example, 1 M for each A and B, our eq'n would be:

rate = [1][1]
1 = 1 x 1
(i've omitted k here for now for simplicity's sake since it would stay the same regardless of what we did to the concentrations)

now, if we doubled both the rates as stated in the question, they would each become 2 M:

rate = 2 x 2
4 = 2 x 2

thus, the reaction rate will change by a factor of 4.

To determine how the reaction rate will change when the concentrations of both A2 and B2 are doubled, we first need to understand the concept of rate law.

The rate law for a reaction specifies how the rate of the reaction depends on the concentrations of the reactants. In this case, the rate law will be of the form:

Rate = k [A2]^x [B2]^y

Where [A2] and [B2] are the concentrations of A2 and B2, respectively, and k is the rate constant. The values of x and y represent the reaction order with respect to A2 and B2, respectively.

Since the reaction is said to proceed by bimolecular collisions between A2 and B2 molecules, we can assume that the reaction is second order overall. This means that x = y = 1.

Therefore, the rate law becomes:

Rate = k [A2] [B2]

Now, if the concentrations of both A2 and B2 are doubled, the new concentrations will be 2[A2] and 2[B2]. Plugging these values into the rate law, we get:

New rate = k (2[A2]) (2[B2]) = 4k [A2] [B2]

Comparing the new rate with the original rate, we can see that the new rate is four times greater than the original rate. Therefore, when the concentrations of both A2 and B2 are doubled, the reaction rate will change by a factor of 4.

Certainly! To determine how the reaction rate will change when the concentrations of A2 and B2 are doubled, we need to consider the overall reaction's rate law, which describes the relationship between the rate of the reaction and the concentrations of the reactants.

The rate law for the given reaction is rate = k[A2]^a[B2]^b, where k is the rate constant and a and b are the reaction orders with respect to A2 and B2, respectively.

In this case, the reaction is said to proceed by bimolecular collisions, which means that the rate of the reaction is directly proportional to the concentrations of both A2 and B2. Therefore, we can assume that a = b = 1.

So, considering the rate law, the reaction rate can be written as rate = k[A2]^1[B2]^1 = k[A2][B2].

Now, let's assume the initial concentrations of A2 and B2 are [A2]₀ and [B2]₀, respectively. When both concentrations are doubled, the new concentrations become 2[A2]₀ and 2[B2]₀.

Substituting the new concentrations into the rate equation, we get the new reaction rate: rate' = k[2A2][2B2] = 4k[A2][B2].

Therefore, the reaction rate will increase by a factor of 4 when the concentrations of both A2 and B2 are doubled.