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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? |
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Answer» 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 |
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