Chemical Kinetics - SS2 Chemistry Past Questions and Answers - page 4
The rate constant for a first-order reaction is found to be 0.025 min^-1. What is the half-life of the reaction?
0.025 min
0.050 min
0.100 min
0.693 min
The rate constant for a second-order reaction is given as 0.002 M-1s-1. What is the unit for the rate constant in a third-order reaction?
M-1s-1
M-2s-1
M-2s
M-3s
The rate law for a reaction is determined to be rate = k[A][B]2. If the concentration of reactant A is tripled while the concentration of reactant B remains unchanged, how does the rate of the reaction change?
The rate remains unchanged.
The rate is tripled.
The rate is doubled.
The rate is halved.
A reaction has a rate constant of 0.002 s^-1. What is the order of the reaction if the concentration of the reactant is raised to the power of 2 in the rate law expression?
Zero order
First order
Second order
Third order
In a zero-order reaction, how does the rate of the reaction change with increasing concentration of the reactant?
The rate remains constant.
The rate decreases.
The rate increases linearly.
The rate increases exponentially.
For the reaction: 2A + 3B -> C, the rate law is given by Rate = k [A]2 [B]. If the initial concentrations of A and B are 0.2 M and 0.3 M, respectively, and the rate constant (k) is 0.01 M-2 s-1, what will be the rate of reaction after 10 seconds?
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.
For a first-order reaction, the rate constant (k) is 0.02 s-1. If the initial concentration of the reactant is 0.8 M, what will be the concentration of the reactant after 30 seconds?
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.
The reaction A -> Products follows pseudo-first-order kinetics with respect to A. In a certain experiment, the concentration of A is 0.1 M, and the observed rate constant is 0.05 s-1. If the half-life of the reaction is 20 seconds, what will be the initial concentration of A in another experiment, where the half-life is 10 seconds?
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.
Given the following rate equation: rate = k[A]2[B], determine the overall reaction order and the individual reaction orders with respect to reactants A and B.
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.
The rate constant (k) for a first-order reaction is 0.025 s-1. If the initial concentration of the reactant is 0.10 M, calculate the concentration after 40 seconds.
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.