Chemical Kinetics - SS2 Chemistry Past Questions and Answers - page 4

Given: [A]₀ = 0.2 M, [B]₀ = 0.3 M, k = 0.01 M^-2 s^-1, t = 10 seconds.

Using the rate law, we can calculate the rate of the reaction:

Rate = k [A]² [B]

Rate = (0.01 M^-2 s^-1) × (0.2 M)² × (0.3 M)

Rate = 0.01 M^-2 s^-1 × 0.04 M² × 0.3 M

Rate = 0.0012 M/s

So, the rate of the reaction after 10 seconds is 0.0012 M/s.

Given: [A]₀ = 0.8 M, k = 0.02 s^-1, t = 30 seconds.

For a first-order reaction, the integrated rate law is given by:

[A] = [A]₀ x e^(-kt)

where [A] is the concentration of the reactant at time t, [A]₀ is the initial concentration, k is the rate constant, and e is the base of the natural logarithm.

Now, let's calculate the concentration of the reactant after 30 seconds:

[A] = 0.8 M x e^{(-0.02 s-1 x 30 s)}

[A] = 0.8 M x e^(-0.6)

[A] ≈ 0.8 M x 0.5488

[A] ≈ 0.439 M

So, the concentration of the reactant after 30 seconds is approximately 0.439 M.

Given: [A]₁ = 0.1 M, k₁ = 0.05 s^-1, t₁/2 = 20 seconds. We need to find [A]₂.

For a pseudo-first-order reaction, the half-life (t₁/2) is given by:

t₁/2 = ln(2) / k

where k is the rate constant.

Now, let's find the rate constant (k₂) for the second experiment where t₁/2 = 10 seconds:

k₂ = ln(2) / t₁/2

k₂ = ln(2) / 10 s

k₂ ≈ 0.0693 s^-1

Now, using the pseudo-first-order rate law, we can find [A]₂:

[A]₂ = k₂ x t₁/2

[A]₂ = 0.0693 s^-1 x 10 s

[A]₂ ≈ 0.693 M

So, the initial concentration of A in the second experiment will be approximately 0.693 M.

The overall reaction order is the sum of the individual reaction orders. From the given rate equation, we can determine the reaction orders as follows:

Overall reaction order = 2 + 1 = 3

The reaction order with respect to reactant A is 2, and the reaction order with respect to reactant B is 1.

For a first-order reaction, the rate equation is given by:

rate = k[A]

We can rearrange the equation to solve for the concentration:

rate = k[A]

[A] = rate / k

Given that the rate constant (k) is 0.025 s^-1 and the initial concentration ([A]₀) is 0.10 M, we can substitute these values into the equation:

[A] = (rate / k) = ([A]₀) x e^(-kt)

[A] = (0.10 M) x e(-0.025 s^-1 x 40 s)

[A] ≈ 0.033 M

Therefore, the concentration of the reactant after 40 seconds is approximately 0.033 M.