Question

The decomposition of dinitrogen pentoxide is described by the chemical equation,
$2{{N}_{2}}{{O}_{5}}\left( g \right)\to 4N{{O}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)$
If the rate of disappearance of ${{N}_{2}}{{O}_{5}}$is equal to the 1.80 mol/min at a particular moment, what is the rate of appearance of $N{{O}_{2}}$ at that moment?
(A) 6.6 mol/min
(B) 1.6 mol/min
(C) 3.6 mol/min
(D) 2.6 mol/min

Answer 1

Hint: The rate of the reaction is described by the chemical kinetics part of the chemistry subject.
The rate depends on the stoichiometric coefficients of the reaction.

Complete step by step solution:
Let us see what we mean by the rate of reaction before solving the given question.
Chemical kinetics is the branch of chemistry where we determine the rate of reaction i.e. by which rate and extent the reaction is taking place in the given conditions.
Rate of reaction is the change in concentration of reactant or product per unit time.
It is expressed as,
\[reac\tan t\to product\]
1. Rate of formation of product = $\dfrac{\Delta product}{\Delta t}$
2. Rate of reaction of reactant = $\dfrac{-\Delta reac\tan t}{\Delta t}$
Negative sign in the reactant part shows that the concentration of the reactant decreases while the reaction proceeds in the forward direction.
If the reaction has stoichiometric coefficients while we balance any equation then the equation for rate of reaction will modify as,
\[a\left( reac\tan t \right)\to b\left( product \right)\]
1. Rate of formation of product = $\dfrac{\Delta product}{b\Delta t}$
2. Rate of reaction of reactant = $\dfrac{-\Delta reac\tan t}{a\Delta t}$
where, a and b are the stoichiometric coefficients of the reactant and product respectively.
Now, let us proceed towards the given illustration,
Given that,
$2{{N}_{2}}{{O}_{5}}\left( g \right)\to 4N{{O}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)$
The decomposition rate of ${{N}_{2}}{{O}_{5}}$ = 1.80 mol/min.
Thus, $-\dfrac{d\left[ {{N}_{2}}{{O}_{5}} \right]}{dt}=1.80mol/\min $
According to the stoichiometry of the given reaction,
$-\dfrac{d\left[ {{N}_{2}}{{O}_{5}} \right]}{2dt}=\dfrac{d\left[ N{{O}_{2}} \right]}{4dt}=\dfrac{d\left[ {{O}_{2}} \right]}{dt}$
As, we need to find the rate of formation of $N{{O}_{2}}$;
$-\dfrac{d\left[ {{N}_{2}}{{O}_{5}} \right]}{2dt}=\dfrac{d\left[ N{{O}_{2}} \right]}{4dt}$
solving the above equation we get,
$\dfrac{d\left[ N{{O}_{2}} \right]}{4dt}$ = $2\times 1.8=3.6mol/\min $
Thus, the rate of appearance or formation of $N{{O}_{2}}$ is 3.60 mol/min.

Therefore option (C) is correct.

Note: Do note to include the stoichiometric coefficients while solving for the rate of the reaction. Making any mistake here will result in chaos while further solving the same.
Negative sign in the reactant side only symbolises it’s significance, it does not make a difference in the calculations.
Check the units of given questions and the options carefully.