
For the elementary reaction $2A\to C$ the concentration of $A$ after $30\text{ minutes}$ was found to be $0.01\text{ mole/liter}$ an if the rate constant of the reaction is $2.5\times {{10}^{-2}}liter\text{ }mole\text{ se}{{\text{c}}^{-1}}$, then the rate of the reaction at $30\text{ minutes}$ is:
Answer
553.8k+ views
Hint: By the rate law we mean the molar concentrations of the reactants raised to the power of their stoichiometric coefficient and depends on the slowest and the rate determining step and can be found by applying the formula as: Rate= $K{{[A]}^{2}}$. Now solve it.
Complete Solution :
First of let’s discuss what is the rate of reaction. Rate of reaction may be defined as the change in any one or the products per unit time and by rate law we mean the rate of reaction in terms of molar concentrations of reactants with each term raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation.
For example, consider the reaction as:
$2{{H}_{2(g)}}+2N{{O}_{(g)}}\to {{N}_{2}}{{O}_{(g)}}+{{H}_{2}}{{O}_{(g)}}$
So, the rate of ${{N}_{2}}O$ in the reaction is as;
$Rate\text{ }of\text{ }{{N}_{2}}O=K{{[{{H}_{2}}]}^{2}}{{[NO]}^{2}}$
- Now considering the statement:
We have been given that:
$2A\to C$
So, the rate of reaction for the given reaction is:
Rate= $K{{[A]}^{2}}$ -------------(1)
Here K is the rate constant
- As we know that:
K=$2.5\times {{10}^{-2}}$ (given)
$[A]=0.01={{10}^{-2}}$ (given)
Then put these value in equation (1), we get:
$\begin{align}
& Rate=2.5\times {{10}^{-2}}\times {{({{10}^{-2}})}^{2}} \\
& \text{ = }2.5\times {{10}^{-6}}\text{ se}{{\text{c}}^{-1}} \\
\end{align}$
Thus, if the concentration of $A$ after $30\text{ minutes}$ was found to be $0.01\text{ mole/liter}$ and if the rate constant of the reaction is $2.5\times {{10}^{-2}}liter\text{ }mole\text{ se}{{\text{c}}^{-1}}$, then the rate of the reaction at $30\text{ minutes}$ is $2.5\times {{10}^{-6}}\text{ se}{{\text{c}}^{-1}}$.
Note: Always keep in mind that the rate of the reaction is always in seconds i.e. in standard units and if the rate constant of the reaction is given in minutes , then first convert it into seconds and then find the rate of the reaction.
Complete Solution :
First of let’s discuss what is the rate of reaction. Rate of reaction may be defined as the change in any one or the products per unit time and by rate law we mean the rate of reaction in terms of molar concentrations of reactants with each term raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation.
For example, consider the reaction as:
$2{{H}_{2(g)}}+2N{{O}_{(g)}}\to {{N}_{2}}{{O}_{(g)}}+{{H}_{2}}{{O}_{(g)}}$
So, the rate of ${{N}_{2}}O$ in the reaction is as;
$Rate\text{ }of\text{ }{{N}_{2}}O=K{{[{{H}_{2}}]}^{2}}{{[NO]}^{2}}$
- Now considering the statement:
We have been given that:
$2A\to C$
So, the rate of reaction for the given reaction is:
Rate= $K{{[A]}^{2}}$ -------------(1)
Here K is the rate constant
- As we know that:
K=$2.5\times {{10}^{-2}}$ (given)
$[A]=0.01={{10}^{-2}}$ (given)
Then put these value in equation (1), we get:
$\begin{align}
& Rate=2.5\times {{10}^{-2}}\times {{({{10}^{-2}})}^{2}} \\
& \text{ = }2.5\times {{10}^{-6}}\text{ se}{{\text{c}}^{-1}} \\
\end{align}$
Thus, if the concentration of $A$ after $30\text{ minutes}$ was found to be $0.01\text{ mole/liter}$ and if the rate constant of the reaction is $2.5\times {{10}^{-2}}liter\text{ }mole\text{ se}{{\text{c}}^{-1}}$, then the rate of the reaction at $30\text{ minutes}$ is $2.5\times {{10}^{-6}}\text{ se}{{\text{c}}^{-1}}$.
Note: Always keep in mind that the rate of the reaction is always in seconds i.e. in standard units and if the rate constant of the reaction is given in minutes , then first convert it into seconds and then find the rate of the reaction.
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