Question

The rate constant for the decomposition of nitrogen dioxide NO2(g) ⟶⟶ NO (g) + 1/2 O2(g) with a laser beam is 1.06 1/M⋅⋅ min. Find the time, in seconds, needed to decrease 2.15 M of NO2 to 1.06 M. Hint: What is the order of the reaction? How can you determine that? Units of k?

Answer 1

Answer:

The time needed to decrease $2.15M$ of $NO_{2}$ to $1.06M$ is $27s.$

Explanation:

Let's find out the order and the unit of the reaction first·

To find out the order of the reaction $:$

 The balanced chemical equation for the decomposition of the nitrogen dioxide is given as,

   $NO_{2}_{(g)}$   →→  $NO_{(g)}\ \ +\ \ \frac{1}{2}O_{2} _{(g)}$

We can rewrite this equation as,

   $2NO_{2}_{(g)}$  →→  $2NO_{(g)}\ \ +\ \ O_{2} _{(g)}$

Order of the reaction can be find out from the rate equation·

     Rate  $=\ K\ [NO_{2} ]^{2}$

∴      order  $=\ \ 2$

So the reaction follows second order kinetics·

To find out the unit of this second order reaction,

     K   $=\ \ \frac{concentration}{time} *\frac{1}{(concentration)^{n} }$

where,

     n  $=$  order of the reaction

     unit of concentration $=\ \ molL^{-1}$

∴    K  $=\ \ \frac{molL^{-1} }{s}\ *\ \frac{1}{(molL^{-1})^{2} }$

     K  $=\ \ mol^{-1} L s^{-1}$

So the unit of the second order reaction is $=\ mol^{-1} L s^{-1}$

To find out the time,

The rate expression for second order reaction is $:$

        $Kt\ =\ \frac{1}{[A_{t}] } \ -\ \frac{1}{[A_{0} ]}$

where,

 K  $=$ rate constant

 t $=$ time

 $[A_{t} ]$ $=$ concentration at t time

 $[A_{0} ]\ =$ Initial concentration

Given that,

 K  $=\ \ 1.061M^{-1} Lmin^{-1}$

 $[A_{t} ]\ =\ 1.06M$

 $[A_{0} ]\ \ =\ \ 2.15M$

On substituting these values

⇒ $Kt\ =\ \frac{1}{1.06}\ -\ \frac{1}{2.15}$

⇒ $Kt\ =\ 0.478$

⇒ $t\ =\ \frac{0.478}{1.061}$

⇒ $t\ =\ 0.45min$

⇒ $t\ =\ 0.45\ *\ 60\ =\ 27s$

Answer 2

Given:

• The balanced chemical equation is given by:
• $NO_{2}(g)$ →→→→ $NO(g)+\frac{1}{2}O_2$
• Rate Constant, K = 1.061  $M^{-1} L min^{-1}$  
• Concentration at time t, $[A_t] = 1.06 M$
• Concentration at time 0, $[A_0] = 2.15 M$  

To Find:

• Order of the reaction
• Time taken to decrease 2.15 M to 1.06 M
• Rate constant (K)

Solution:

• First, we should find the order of the reaction.
• Depending on the order of the reaction we get to know the units used here.
• To find order we are rearranging the balanced equation as:
• $2NO_{2}(g)$  →→→→ $2NO(g) + O_2$  
• By determining the rate of the reaction, we can find the order of the reaction:
• Rate = $K[NO_{2}]^{2}$  
• Therefore, Order = 2 (It is a second order reaction)
• Unit of second order reaction is K = $\frac{concentration}{time}*\frac{1}{(concentration)^{n} }$
• Unit of concentration = mol/L
• n = order
• Substituting the values we get, K = $\frac{molL^{-1}}{s} *\frac{1}{(molL^{-1})^2} = mol^{-1}Ls^{-1}$
• The rate expression for second order reaction is given by:
• $Kt = \frac{1}{[A_t]} -\frac{1}{[A_0]}$
• Substitute the given values.
• To find t:
• Kt = $\frac{1}{1.06}-\frac{1}{2.15}$
• Kt = $\frac{1.09}{2.279} = 0.478$  
• t = 0.478/K = 0.478/1.061 = 0.45 min
• t = 0.45*60 = 27 secs.

Time taken to decrease from 2.15M to 1.06 is 27 secs.