The following data were obtained during the first order thermal decomposition of `N_2O_5 `(g) at constant volume:
` 2N_2 O_5 (g) to 2N_2 O_4 (g) +O_2 (g)`

Answer

`k= 4.98 xx 10^(-4) s^(-1)

Explanation

`2N_2 O_5 to 2N_2 O_4 (g) +O_(2) (g) `
Let the pressure of `N_2 O_5` decrease by 2p atm. Since 2 moles of `N_2 O_5` decompose to give two moles of `N_2 O_4` and one mole of `O_2 (g)`, increase in pressure of `N_2 O_4=2p` and that of `O_2(g)= `p atm.
` 2N_2 O_5 (g) to 2N_2 O_(4) (g) +O_(2) (g) P_t =p_(N_2 O_4) +PO_2`
`(i) At t=0 s (0.5 - 2p ) atm 2 p atm p atm `
` p_t = ( 0.5 - 2 p ) + 2p + p = 0.5 + p`
` p=p_t - 0.5`
` P_(N_2 O_5) = 0.5 - 2p = 0.5 - 2 (P_t - 0.5 ) = 1.5 - 2pt.`
(ii) At ` t=100 s , p_t = 0.512 atm `
` p_(N_2 O_5 ) = 1.5 -2 xx 0.512 = 0.476 atm`
`p_(N_2 O_5) = 1.5 - 2 xx 0.512 = 0.476 atm `
Now ` k=( 2.303)/(t) log"" (p_i)/(p_A) = (2.303 )/(100 s ) log"" ( 0.5 atm ) /( 0.476 )`
`= ( 2.303 )/( 100 s ) xx 0.0216 = 4.98 xx 10^(-4) s^(-1)`