
Initial Concentration Expt. No. $\left[ NO \right]$ $\left[ C{{l}_{2}} \right]$ Initial rate 1. 0.010 0.010 $1.2\times {{10}^{-4}}$ 2. 0.010 0.020 $2.4\times {{10}^{-4}}$ 3. 0.020 0.020 $9.6\times {{10}^{-4}}$
For the reaction, $2NO+C{{l}_{2}}\to 2NOCl$, at 300K, following data is obtained:
Write rate law and order of the reaction? Also, calculate the specific rate constant.
Initial Concentration | |||
Expt. No. | $\left[ NO \right]$ | $\left[ C{{l}_{2}} \right]$ | Initial rate |
1. | 0.010 | 0.010 | $1.2\times {{10}^{-4}}$ |
2. | 0.010 | 0.020 | $2.4\times {{10}^{-4}}$ |
3. | 0.020 | 0.020 | $9.6\times {{10}^{-4}}$ |
Answer
510.3k+ views
Hint: The rate of a chemical reaction is the change in the concentration of reactants or products per unit time. Think about the relation between the rate and molar concentration of reacting species. Take a look at the number of moles of each species present in the reaction.
Complete answer:
-The rate law is defined as an experimentally determined equation that gives the rate of the reaction in terms of molar concentrations of the reactants. The rate of a reaction is directly proportional to the concentration of the reactants at any given point of time.
- Consider a reaction, $\text{aA + bB }\to \text{Products}$
\[\begin{align}
& rate\propto {{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}} \\
& rate=k{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}} \\
\end{align}\]
where k is the rate constant or specific reaction rate constant. The rate constant is the rate of the reaction if all the concentrations of reactants are equal to unity.
Now let’s see for the given reaction, $2NO+C{{l}_{2}}\to 2NOCl$
Rate law is given as, $rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
Order of the reaction is experimentally determined and can be given as a sum of the exponential terms of the concentrations of the reactants in a rate law. So, the order of the given reaction is $2+1=3$. Therefore, the given reaction is a third order reaction.
Now let’s calculate the rate constant of the reaction.
For first experiment,
$\left[ NO \right]=0.010M$, $\left[ C{{l}_{2}} \right]=0.010M$ and rate =$1.2\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{1.2\times {{10}^{-4}}}{{{(0.010)}^{2}}\times 0.010}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
Similarly, for second experiment,
$\left[ NO \right]=0.010M$, $\left[ C{{l}_{2}} \right]=0.020M$ and rate =$2.4\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{2.4\times {{10}^{-4}}}{{{(0.010)}^{2}}\times 0.020}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
In the same way, for third experiment,
$\left[ NO \right]=0.020M$, $\left[ C{{l}_{2}} \right]=0.020M$ and rate =$9.6\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{9.6\times {{10}^{-4}}}{{{(0.020)}^{2}}\times 0.020}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
Thus, the specific rate constant for the reaction, $2NO+C{{l}_{2}}\to 2NOCl$, at 300K, is $1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$.
Note:
Don’t get confused with the term rate constant and specific rate constant. Both the terminologies are used to refer to the same thing. Remember that the rate of a reaction at a given temperature depends only on the molar concentration of the reactants and not on the molar concentration of products of the reaction.
Complete answer:
-The rate law is defined as an experimentally determined equation that gives the rate of the reaction in terms of molar concentrations of the reactants. The rate of a reaction is directly proportional to the concentration of the reactants at any given point of time.
- Consider a reaction, $\text{aA + bB }\to \text{Products}$
\[\begin{align}
& rate\propto {{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}} \\
& rate=k{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}} \\
\end{align}\]
where k is the rate constant or specific reaction rate constant. The rate constant is the rate of the reaction if all the concentrations of reactants are equal to unity.
Now let’s see for the given reaction, $2NO+C{{l}_{2}}\to 2NOCl$
Rate law is given as, $rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
Order of the reaction is experimentally determined and can be given as a sum of the exponential terms of the concentrations of the reactants in a rate law. So, the order of the given reaction is $2+1=3$. Therefore, the given reaction is a third order reaction.
Now let’s calculate the rate constant of the reaction.
For first experiment,
$\left[ NO \right]=0.010M$, $\left[ C{{l}_{2}} \right]=0.010M$ and rate =$1.2\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{1.2\times {{10}^{-4}}}{{{(0.010)}^{2}}\times 0.010}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
Similarly, for second experiment,
$\left[ NO \right]=0.010M$, $\left[ C{{l}_{2}} \right]=0.020M$ and rate =$2.4\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{2.4\times {{10}^{-4}}}{{{(0.010)}^{2}}\times 0.020}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
In the same way, for third experiment,
$\left[ NO \right]=0.020M$, $\left[ C{{l}_{2}} \right]=0.020M$ and rate =$9.6\times {{10}^{-4}}M/s$
$rate=k{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]$
\[k=\dfrac{rate}{{{\left[ NO \right]}^{2}}\left[ C{{l}_{2}} \right]}=\dfrac{9.6\times {{10}^{-4}}}{{{(0.020)}^{2}}\times 0.020}=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}\]
Therefore, specific rate constant for first experiment, $k=1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$
Thus, the specific rate constant for the reaction, $2NO+C{{l}_{2}}\to 2NOCl$, at 300K, is $1.2\times {{10}^{-2}}{{M}^{-2}}{{s}^{-1}}$.
Note:
Don’t get confused with the term rate constant and specific rate constant. Both the terminologies are used to refer to the same thing. Remember that the rate of a reaction at a given temperature depends only on the molar concentration of the reactants and not on the molar concentration of products of the reaction.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

The non protein part of an enzyme is a A Prosthetic class 11 biology CBSE

Which of the following blood vessels in the circulatory class 11 biology CBSE

What is a zygomorphic flower Give example class 11 biology CBSE

The deoxygenated blood from the hind limbs of the frog class 11 biology CBSE

What is the function of copulatory pads in the forelimbs class 11 biology CBSE
