With rate constant of `5 xx 10^(-4)sec^(-1)` at `45^(@)`C, If initial concentration of `N_(2)O_(5)` is 0.25 M, Calcualte the concentration after 2 min
With rate constant of `5 xx 10^(-4)sec^(-1)` at `45^(@)`C, If initial concentration of `N_(2)O_(5)` is 0.25 M, Calcualte the concentration after 2 minutes. Also calculate half life for the decomposition of `N_(2)O_(5)` ?
`2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g)`
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For a first order reaction, `t=2.303/k log a/(a-x)`
a=0.25 M, `t=2xx 60` =120s, k`=5 xx 10^(-4)s^(-1)`
`log a/(a-x) = (k xx t)/2.303 = (5 xx 10^(-4)s^(-1) xx (120s))/(2.303)= 0.02605`
`a/(a-x) = "Antilog" 0.02605 = 1.0618`
Concentration after 2 minutes i.e. (a-x) = `a/1.0618 = (0.25M)/(1.6018 = 0.2354 M`
Half life period `(t_(1/2))= 0.693/k=0.693/(5 xx 10^(-4)s^(-1))`=1386 s
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