Question

Given, \( \left.2 NO _{2}+ O _{2}( g )- NOQ _{ Cg }\right)^{ h } \) rate \( =k\left[ NO ^{2}\right]^{2}\left[ O _{2}\right]^{3} \). By how many times does the rate of the reaction change when the value of the reaction vessel is VL and reduced to 1/3rd of its original volume? Will there be any change in the order \( 0 . b \) the reaction?

Answer 1

2NO₂(g) + O₂(g) → product

Given,

rate of reaction = k[NO₂]² [O₂]³

We know that for a gaseous reaction, concentration of gaseous species is directly proportional to the partial pressure of that species. So on decreasing the volume of reaction vessels, the partial pressure of both NO₂ and O₂ increases and rate of reaction increases

Let say initial volume of reaction vessel is V and number of moles of NO₂ and O₂ are a and b respectively.

Then, 

initial rate of reaction  \(r_1 = k \left(\frac aV\right)^2 \left(\frac bV\right)^3\)     ......(i)

On decreasing the volume of vessel to \(\frac13rd\) of initial,

The number of moles of NO₂ and O₂ are remain same but concentrations are changed.

Concentration of NO₂ = \(\frac{a}{\frac V3} = \frac {3a}V\)

Concentration of O₂ = \(\frac{b}{\frac V3} = \frac {3b}V\) 

rate of reaction after decreasing the volume \(r_2= k \left(\frac {3a}V\right)^2 \left(\frac {3b}V\right)^3\)   ......(ii)

From equation (i) and (ii)

\(\frac{r_2}{r_1} = \cfrac{(\frac{3a}{V})^2}{(\frac a V)^2} \times \cfrac{(\frac{3b}{V})^3}{(\frac bV)^3}\)

\(\frac{r_2}{r_1} = 3^2 \times 3^3\)

\(\frac{r_2}{r_1} = 3^5\)

\(r_2= 3^5r_1\)

\(r_2= 243r_1\)

Hence, the rate of reaction increases by 243 time of the initial rate.

There is no change in the order of reaction order of reaction will be (2 + 3)  = 5