For a reaction $A + B \to \text{Products}$,the rate law is: $\text{Rate} = k[A][B]^{3/2}$. Can the reaction be an elementary reaction? Explain.
(N/A) An elementary reaction is a single-step process where the order of the reaction is equal to the sum of the stoichiometric coefficients of the reactants.
For an elementary reaction,the order must be an integer (usually $1, 2, \text{or } 3$).
In the given rate law,$\text{Rate} = k[A]^1[B]^{3/2}$,the overall order of the reaction is $1 + 3/2 = 5/2 = 2.5$.
Since the order of the reaction is fractional,it cannot be an elementary reaction.
For an elementary reaction,the order must be an integer (usually $1, 2, \text{or } 3$).
In the given rate law,$\text{Rate} = k[A]^1[B]^{3/2}$,the overall order of the reaction is $1 + 3/2 = 5/2 = 2.5$.
Since the order of the reaction is fractional,it cannot be an elementary reaction.
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Similar Questions
Consider the following two reactions:
$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$
$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$
The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as:
$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$
$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$
The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as:
Medium
View SolutionFor the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:
If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?
Medium
View SolutionFor the reaction $Cl_{2(aq)} + H_2S_{(aq)} \to S_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)}$, the rate law is given by $\text{Rate} = K[Cl_2][H_2S]$. Which of the following mechanisms is consistent with this rate law?
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)
Medium
View SolutionWhy is the probability of a reaction with molecularity higher than $3$ very rare?
Easy
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