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.A. `3.39xx10^(-4)S^(-1)`B. `1.39xx10^(-5)S^(-1)`C. `5.45xx10^(-4)S^(-1)`D. `4.91xx10^(-4)S^(-1)`

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.A. `3.39xx10^(-4)S^(-1)`B. `1.39xx10^(-5)S^(-1)`C. `5.45xx10^(-4)S^(-1)`D. `4.91xx10^(-4)S^(-1)`
Answer» When t=0 `" "` 0.5 atm At time t `(0.5-2x)""atm" " "2xatm " "xatm"` Total pressure `(p_(t))=P_(N2O5)+ p_(N_2O_5)+P_(O2)` `=0.5-2x+2x+x` `p_(t)=0.5+x` `x=p_(t)-0.5` `p_(N_2O_5)=0.5-2x` `=0.5-2(p_(t)-0.5)=0.5-2p_(t)+1` `p_(N_2O_5)=1.5-2p_(t)` At `t=100_(s),p_(t)=0.512"atm"` `p_(N_2O_5)=1.5-2xx0.512` `=1.5-1.024=0476"atm"` `k=(2.303)/(t)"log"[(p_(0)(N_(2)O_5))/(p_(A)(N_(2)O_(5)))]` `=(2.303)/(100)"log"(0.5)/(0.476)=(2.303)/(100)xx0.02136` `=4.92xx10^(-4)S^(-1)`

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