
The unit of rate constant for reactions of second order is:
A. $Lmo{{l}^{-1}}{{s}^{-1}}$
B. ${{L}^{-1}}mol{{s}^{-1}}$
C. $Lmol{{s}^{-1}}$
D. ${{s}^{-1}}$
Answer
566.1k+ views
Hint: The unit for rate constant depends on the rate law as that links the rate with the concentration. Different order reactions have different rate laws and thus different units for the expressions of their rate law.
Complete answer:
- The rate law can be shown as
\[rate=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}\]
where A and B are reactants of the equation and k is a constant parameter. The sum of x and y gives the overall order of the reaction.
-If the rate of the equation does not depend on the reactant concentration; i.e; it is constant irrespective of the reactant concentration, then the equation is zero order.
-If the rate depends on one reactant, it is a first order reaction and if it depends on two reactants or square of 1 reactant, then it is a second order reaction.
- The equation for first order reaction is $\ln \left[ A \right]=-kt+\ln \left[ {{A}_{0}} \right]$
The rate law for second order reaction is $\dfrac{1}{\left[ A \right]}=\dfrac{1}{{{\left[ A \right]}_{\circ }}}+kt$
The rate law for nth order reaction is $\dfrac{1}{{{\left[ A \right]}^{n-1}}}=\dfrac{1}{{{\left[ A \right]}_{\circ }}^{n-1}}+\left( n-1 \right)kt$
-Order of a reaction is very crucial as it tells us the method in which the reaction will occur and the products formed by such method. Same reactants will give different products if they have different orders as they will occur in completely different forms.
-The rate law for the second order is shown as $\dfrac{1}{\left[ A \right]}=\dfrac{1}{{{\left[ A \right]}_{\circ }}}+kt$
So the equation for rate can be given as rate = k${{\left[ A \right]}^{2}}$
Solving for the units of k we get
$\begin{align}
& mol{{L}^{-1}}{{s}^{-1}}=k{{\left( mol{{L}^{-1}} \right)}^{2}} \\
& \Rightarrow \text{k=Lmo}{{\text{l}}^{-1}}{{s}^{-1}} \\
\end{align}$
Therefore the correct option is A.
Note:
The order can also be found by knowing the time duration in which the reaction is being completed. The duration in which half of the reaction gets completed is called half-life of the reaction and is denoted by ${{t}_{1/2}}$.
Complete answer:
- The rate law can be shown as
\[rate=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}\]
where A and B are reactants of the equation and k is a constant parameter. The sum of x and y gives the overall order of the reaction.
-If the rate of the equation does not depend on the reactant concentration; i.e; it is constant irrespective of the reactant concentration, then the equation is zero order.
-If the rate depends on one reactant, it is a first order reaction and if it depends on two reactants or square of 1 reactant, then it is a second order reaction.
- The equation for first order reaction is $\ln \left[ A \right]=-kt+\ln \left[ {{A}_{0}} \right]$
The rate law for second order reaction is $\dfrac{1}{\left[ A \right]}=\dfrac{1}{{{\left[ A \right]}_{\circ }}}+kt$
The rate law for nth order reaction is $\dfrac{1}{{{\left[ A \right]}^{n-1}}}=\dfrac{1}{{{\left[ A \right]}_{\circ }}^{n-1}}+\left( n-1 \right)kt$
-Order of a reaction is very crucial as it tells us the method in which the reaction will occur and the products formed by such method. Same reactants will give different products if they have different orders as they will occur in completely different forms.
-The rate law for the second order is shown as $\dfrac{1}{\left[ A \right]}=\dfrac{1}{{{\left[ A \right]}_{\circ }}}+kt$
So the equation for rate can be given as rate = k${{\left[ A \right]}^{2}}$
Solving for the units of k we get
$\begin{align}
& mol{{L}^{-1}}{{s}^{-1}}=k{{\left( mol{{L}^{-1}} \right)}^{2}} \\
& \Rightarrow \text{k=Lmo}{{\text{l}}^{-1}}{{s}^{-1}} \\
\end{align}$
Therefore the correct option is A.
Note:
The order can also be found by knowing the time duration in which the reaction is being completed. The duration in which half of the reaction gets completed is called half-life of the reaction and is denoted by ${{t}_{1/2}}$.
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