The following data were obtained during the first thermal decompoistion of `N_(2)O_(5) (g)` at constant volume. `2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)` `|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}|` Calculate the rate constant.

The following data were obtained during the first thermal decompoistio

1.	The following data were obtained during the first thermal decompoistion of `N_(2)O_(5) (g)` at constant volume. `2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)` `\|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}\|` Calculate the rate constant.
Answer» Let the pressure of `N_(2)O_(5)` (g) decrease by 2x atm . As two moles of `N_(2)O_(5)` decompose to give two moles of `N_(2)O_(4)` (g) and one mole of `O_(2)` (g), the pressure of `N_(2)O_(4)` (g) increases by 2x atm and that of `O_(2)`(g) increases by x atm. `2N_(2)O_(5) (g) to 2N_(2)O_(4)(g) + O_(2) (g)` `{:("Start "t=0," "0.5" atm",0 " atm",0" atm"),("At time t",(0.5-2x)" atm",2x" atm",x" atm"):}` `p_(t)=P_(N_(2)O_(5)+P_(O_(2)))` =`(0.5-2x)+2x+x=0.5+x` `x=P_(t)-0.5` `P_(N_(2)O_(5) =0.5-2x` `=0.5-2(P_(1)-0.5)=1.5-2p_(t)` At `t=100S,P_(1)=0.512` atm `P_(N_(2)O_(5) =1.5-2xx 0.512=0.476 " atm" ` Using equation (4.16) `k=2.303/t"log"P_(i)/P_(A)=2.303/(100s)"log"(0.5atm)/(0.476atm)` = `2.303/(100s)xx0.0216=4.98xx10^(-4)S^(-1)`

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The following data were obtained during the first thermal decompoistion of `N_(2)O_(5) (g)` at constant volume. `2N_(2)O_(5) (g) rarr 2N_(2)O_(4) (g)+O_(2)(g)` `|{:("S.No.","Time (s)","Total pressure (atm)"),(i.,0,0.5),(ii.,100,0.512):}|` Calculate the rate constant.

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