Test: Zero, First, Second and Third Order reactions - Chemistry MCQ

To calculate the time taken for 87.5% completion of a first-order reaction, we can use the following concepts:

For a first-order reaction, the time required to complete a certain percentage of the reaction is given by the first-order rate equation:

where:

• ttt is the time taken for the reaction to reach a certain completion.
• kkk is the rate constant.
• [R]0[R]_0[R]0​ is the initial concentration of the reactant.
• [R][R][R] is the concentration of the reactant at time ttt.

The rate constant kkk for a first-order reaction can be determined from the half-life using the relation:

The other solution can be:-

Reaction is 50 percent complete in 30 minutes. Hence, t1/2 = 30 minutes
75 percent of the reaction is completed in two half-lives. Hence, t = 2 × 30 = 60 minutes
87.5 percent of the reaction is completed in three half-lives. Hence, t = 3 × 30 = 90 minutes.

Given, rate constant k= 0.00058 s^-1
Time t= 50 minutes = 50 × 60 = 3000 seconds
First-order integrated rate equation, k = 

a=82.44 percent.

Given, the reaction is 70 percent complete so, a = 70
Time t = 20 minutes
First order integrated rate equation, k = 

The time taken for the zero-order reaction to complete can be calculated as follows:
When the reaction is complete, [A]0 = 0
Therefore, k = [A]₀/t or t_100% = [A]₀/k.

Given,
Arrhenius factor(A) = 4 x 10¹³ collisions/sec
Activation energy (E_a)=98.6KJ/mol=98.6 x 10³J/mol, T=303 K
log K = log A – (E_a/2.303RT)
log K = log (4 x 10¹³) – (98.6 x 10³)/(2.303 x 8.314 x 303)
log K = 13.6020 – (98.6 x 103/5801.584)
log k = -3.39
K = 10^-3.39
K = 4.07 x 10^-4.

t_25% = (1-0.75)/K
K = 0.25/30
t_50% = (1 x 30)/(2 x 0.25)
t_50% = 15/0.25
t_50% = 60s.

Given,
K = 0.01s^-1
t1/2 = 0.693/0.01
t1/2 = 69.3s
[R] = [R]₀/2ⁿ
2ⁿ = [R]₀/[R]
2ⁿ = 2.4/0.3
2ⁿ = 8
n = 3 (number of half-lives)
For 1 half-life t_1/2 = 69.3s
For 3 half-life 3t_1/2 = 3 x 69.3s = 207.9s.

Given, the reaction is 75 percent complete so, a = 75
We know, t = 

t₇₅ = 2 × t₅₀.

Given, the reaction is 75 complete so, a=75
Rate of the reaction k=0.023 min-1
First-order integrated rate equation, k = 

Half-life (t1⁄2) is the time required for a quantity to reduce to half of its initial value. The half-life of a zero-order reaction is directly proportional to its initial concentration. They are related as:
t_1/2 = [R]₀/2k.

Given,
N₂O₅ → 2NO₂ + 1/2 O₂
Rate = k[N₂O₅] ¹
Rate =6.25 x 10^-4 x [1.25]
Rate = 7.75 x 10^-4.

A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. The integrated rate equation for a first order reaction in exponential form is [A] = [A]₀e^-kt.

Given,
K=7.5 x 10^-3 s^-1
K = (2.303/t) x log([R]₀/[R])
t = (2.303/7.5 x 10^-3) x log([100]/[25])
t = (2.303/7.5 x 10^-3) x 0.6
t = 38.4s.

The unit of the rate constant k for a zero-order reaction is mol L^-1 s^-1. For a first-order reaction, the unit of k is s^-1. The unit of the rate constant k for a second-order reaction is mol^-1 L s^-1.

The rate constant is defined as the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a chemical reaction. It is denoted by k.
For a first-order reaction, the unit of k is s^-1.