Question

The rate of the reaction $\text{ 2}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\to \text{ 4N}{{\text{O}}_{\text{2}}}\text{ + }{{\text{O}}_{\text{2}}}\text{ }$ can be written in three ways:
$\text{ }\begin{matrix}
   -\dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
   \dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k }\!\!'\!\!\text{ }\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
   \dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k''}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
\end{matrix}\text{ }$
Relationship between $\text{ k }$ and $\text{ k }\!\!'\!\!\text{ }$ between $\text{ k }$ and $\text{ k'' }$ are:
A) $\text{ k }\!\!'\!\!\text{ = 2k ; k'' = k }$
B) $\text{ k }\!\!'\!\!\text{ = 2k ; k'' = }\dfrac{\text{k}}{2}\text{ }$
C) $\text{ k }\!\!'\!\!\text{ = 2k ; k'' = 2k }$
D) $\text{ k }\!\!'\!\!\text{ = k ; k'' = k }$

Answer 1

Hint: For a chemical reaction, the rate of reaction at any time will depend on the concentration of reactant. As reaction proceeds the concentration of reactant keeps on falling. For a general reaction of $\text{ aA + bB }\to \text{ cC + dD }$
The rate of reaction is repressed in terms of decreases in the concentration of reactant per mole or the increase in the concentration of product per mole. Accordingly,
$\text{ r = }-\dfrac{1}{a}\dfrac{d{{C}_{A}}}{dt}=-\dfrac{1}{b}\dfrac{d{{C}_{B}}}{dt}=\dfrac{1}{c}\dfrac{d{{C}_{C}}}{dt}=\dfrac{1}{d}\dfrac{d{{C}_{D}}}{dt}\text{ }$

Complete step by step answer:
Here, we have given that the nitrogen pentoxide \[\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\] decomposes into the nitrogen dioxide \[\text{ N}{{\text{O}}_{2}}\text{ }\]and oxygen gas \[\text{ }{{\text{O}}_{2}}\text{ }\]. The reaction is as shown below,
$\text{ 2}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\to \text{ 4N}{{\text{O}}_{\text{2}}}\text{ + }{{\text{O}}_{\text{2}}}\text{ }$
The rate of the reaction of decomposition of the nitrogen pentoxide can be written in terms of the decrease in the concentration of reactant which is \[\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\] or in terms of the decrease in the concentration of \[\text{ N}{{\text{O}}_{2}}\text{ }\] and \[\text{ }{{\text{O}}_{2}}\text{ }\].the rate of reaction is expressed as the follows,
$\text{ rate = }-\dfrac{1}{2}\dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt}\text{ = }\dfrac{1}{4}\dfrac{d(\text{N}{{\text{O}}_{\text{2}}})}{dt}\text{ }=\dfrac{d({{\text{O}}_{\text{2}}})}{dt}\text{ }$ (1)
We know that the,
$\text{ }\begin{matrix}
   -\dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
   \dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k }\!\!'\!\!\text{ }\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
   \dfrac{d({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}})}{dt} & = & \text{k''}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right] \\
\end{matrix}\text{ }$ (2)
On substituting values in the equation (2) in (1).The rate equations can be written as,
$\text{ rate = }\dfrac{1}{2}\text{k}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right]\text{ = }\dfrac{1}{4}\text{k }\!\!'\!\!\text{ }\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right]\text{ }=\text{k''}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right]\text{ }$

On solving further cancel out the \[\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}} \right]\] form the rate equation, we get,
$\text{ rate = }\dfrac{\text{k}}{\text{2}}\text{ = }\dfrac{\text{k }\!\!'\!\!\text{ }}{\text{4}}\text{ = k'' }$
We are interested to find out the relationship between $\text{ k }$ and $\text{ k }\!\!'\!\!\text{ }$ between $\text{ k }$ and $\text{ k'' }$. On solving further we get the relations as follows,
$\begin{align}
 & \text{ }\dfrac{\text{k}}{\text{2}}\text{ = }\dfrac{\text{k }\!\!'\!\!\text{ }}{\text{4}}\text{ } \\
 & \therefore \text{ k }\!\!'\!\!\text{ = 2k } \\
\end{align}$
Similarly,
$\text{ }\dfrac{\text{k}}{\text{2}}\text{ = k'' }$
Thus, relationship between $\text{ k }$ and $\text{ k }\!\!'\!\!\text{ }$ between $\text{ k }$ and $\text{ k'' }$ is, \[\text{k }\!\!'\!\!\text{ = 2k }\] and \[\text{ }\dfrac{\text{k}}{\text{2}}\text{ = k'' }\].
So, the correct answer is “Option B”.

Note: Note that, the \[\text{ }-d{{C}_{A}}\text{ or }-d{{C}_{{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}}}\text{ }\] is an infinitesimally small decrease in the concentration of reactant in a small interval of time $\text{ }dt\text{ }$.the k is also referred as the velocity of a reaction. The concentration is expressed in terms of molarity. Here, the 2 moles of \[\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{ }\]decomposes and produces 4 moles of \[\text{ N}{{\text{O}}_{2}}\text{ }\] and 1 mole of \[\text{ }{{\text{O}}_{2}}\text{ }\],thus the rate of disappearance of reactants and appearance of product may be different, but are related.