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Methods of Determining Reaction Order - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

9 Questions around this concept.

Solve by difficulty

For a reaction A + B → products, the rate of reaction was doubled when concentration of A was doubled. When concentration of A and B both was doubled, the rate was again doubled, order of reaction with respect to A and B are respectively -

1, 1

2, 0

1, 0

0, 1

View Solution

Consider the reaction :

$
\mathrm{C}_{2(a q)}+\mathrm{H}_2 \mathrm{~S}_{(a q)} \longrightarrow \mathrm{S}_{(5)}+2 \mathrm{H}_{(2 q)}^{+}+2 \mathrm{C}_{(2 q)}^{-}
$

The rate of reaction for this reaction is

$
\text { Rate }=K\left[\mathrm{Cl}_2\right]\left[\mathrm{H}_2 \mathrm{~S}\right]
$

Which of these mechanism is/are consistent with this rate equation ?
A $\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{Cl}^{+}+\mathrm{HS}^{-}$(slow)

$
\mathrm{Cl}^{+}+\mathrm{HS}^{-} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+S(\text { fast })
$

B $H_2 S \Leftrightarrow H^{+}+H S^{-}$(fast equilibrium $)$

$
\mathrm{Cl}_2+\mathrm{HS}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{H}^{+}+\mathrm{S}(\text { Slow })
$

A only

B only

Both A and B

Neither A nor B

View Solution

Concepts Covered - 3

How to Determine Order of Reaction: Half Life Method

It is used when the rate law involves only one concentration term.

$\mathrm{t}_{1 / 2} \propto(\mathrm{a})^{1-\mathrm{n}}$
or $\mathrm{t}_{1 / 2} \propto 1 / \mathrm{a}^{\mathrm{n}-1}$

For two different concentrations, we have:

$\frac{\left(\mathrm{t}^{1 / 2}\right)_1}{\left(\mathrm{t}^{1 / 2}\right)_2}=\left(\frac{\mathrm{a}_2}{\mathrm{a}_1}\right)^{\mathrm{n}-1}$

On taking logarithms on both sides, we get:

$\log _{10} \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=(\mathrm{n}-1) \log _{10}\left(\mathrm{a}_2 / \mathrm{a}_1\right)$

Hence,

$\mathrm{n}=1+\frac{\log \left(\mathrm{t}^{1 / 2}\right)_1-\log \left(\mathrm{t}^{1 / 2}\right)_2}{\log \mathrm{a}_2-\log \mathrm{a}_1}$
Here, n is the order of the reaction.

How to Determine Order of Reaction: Graphical Method

Here graphs are plotted between rate and concentration to find the order of the reaction.

$\left[\right.$ Rate $\left.=\mathrm{k}(\text { concentration })^{\mathrm{n}}\right]$

Plots of Rate vs Concentration

How to Determine Order of Reaction - Integrated Rate Law Method

If the data for time(t) and [A] is given then this method is applicable. Thus follows the steps given below to find the order of reaction by using the integrated rate law method.

Check for First Order:
1. Use the formula given below to find out the two values of k as k₁and k₂.$\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10}\left[\frac{\mathrm{~A}_{\mathrm{o}}}{\mathrm{A}}\right]$
2. If these two values k₁and k₂ are same, then this given reaction is of first-order. But if k₁≠ k₂, then check for zero-order.
Check for Zero-Order:
1. Use the formula given below to find out the two values of k as k₁and k₂.$\mathrm{k}=\frac{\mathrm{A}_{\mathrm{o}}-\mathrm{A}}{\mathrm{t}}$
2. Again, if these two values k₁and k₂ are same, then this given reaction is of zero-order. But if k₁≠ k₂, then check for second-order.
Check for Third-Order:
1. Use the formula given below to find out the two values of k as k₁and k₂.$\mathrm{k}=\frac{1}{\mathrm{t}}\left[\frac{1}{\mathrm{~A}}-\frac{1}{\mathrm{~A}_{\mathrm{o}}}\right]$
2. Further, if these two values k₁and k₂ are same, then this given reaction is of second-order. But if k₁≠ k₂, then check for third-order and so on.