Question

For the decomposition reaction of \[{N_2}{O_5}\], the rate of formation of \[N{O_2}\] is 3.5 mol/Ls. Calculate the rate of disappearance of \[{N_2}{O_5}\]

Answer 1

Hint: To solve this question, we need to first understand the definition and application of the Rate law, which is used to find the rate of a reaction. Then, we need to write the chemical equation of the reaction given to us, and form a corresponding rate law. On the basis of this equation and the data given to us, we can find the final answer.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Rate law can be understood as an equation which helps us in finding the rate of a given chemical equation, on the basis of the concentrations of the reactants, and the reaction order of each reactant in the given chemical equation. To put it in simpler terms, the rate law is a simple mathematical tool which helps us in understanding the chemical kinetics of any chemical reaction. The rate law can be represented in a mathematical format as follows:
Rate \[ = k{[A]^x}{[B]^y}\]
Where k is the rate constant, [A] and [B] are the concentrations of the reactants and x and y are the rate orders of the individual reactants.
The chemical reaction for the decomposition reaction of \[{N_2}{O_5}\] can be given as:
 \[2{N_2}{O_5} \to 4N{O_2} + {O_2}\]
Hence, the rate of decomposition of dinitrogen pentoxide is given as \[ - \dfrac{{\Delta [{N_2}{O_5}]}}{{\Delta t}}\]
We can observe that the decomposition of 2 moles of \[{N_2}{O_5}\] results in the formation of 4 moles of \[N{O_2}\]. Hence, these two quantities can be related as:
\[ (2) \left( { - \dfrac{{\Delta [{N_2}{O_5}]}}{{\Delta t}}} \right) = (4)\left( {\dfrac{{\Delta [N{O_2}]}}{{\Delta t}}} \right)\]
$\Rightarrow$ \[\left( { - \dfrac{{\Delta [{N_2}{O_5}]}}{{\Delta t}}} \right) = 2\left( {\dfrac{{\Delta [N{O_2}]}}{{\Delta t}}} \right)\]
$\Rightarrow$ \[\left( { - \dfrac{{\Delta [{N_2}{O_5}]}}{{\Delta t}}} \right) = 2(3.5)\]
$\Rightarrow$ \[\left( { - \dfrac{{\Delta [{N_2}{O_5}]}}{{\Delta t}}} \right) = 7\dfrac{{mol}}{{Ls}}\]

Hence, the rate of disappearance of \[{N_2}{O_5}\] is 7 mol/Ls

Note: Knowing the rate law, an expression relating the rate to the concentrations of reactants, can help a chemist adjust the reaction conditions to get a more suitable rate.