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With rate constant of 5 xx 10^(-4)sec^(-...

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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The decomposition of "N"_(2)"O"_(5)(g), i.e.,"N"_(2)"O"_(5)(g)to4"NO"_(2)(g)+"O"_(2)(g) is a first order reaction with a rate constant of 5xx10^(-4)"sec"^(-1)" at "45^(@)C . If intial concentration of "N"_(2)"O"_(5) is 0.25 M, calculate its concentration after 2 min. Also calculate half life for the decomposition of "N"_(2)"O"_(5)(g).

For the first order reaction 2N_(2)O_(5)(g) rarr 4NO_(2)(g) + O_(2)(g)

For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

Using the concentration time equation for a first order reaction : The decomposition of N_(2)O_(5) to NO_(2) and O_(2) is first order, with a rate constant of 4.80xx10^(-4)//s att 45^(@)C N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g) (a) If the initial concentration of N_(2)O_(5) is 1.65xx10^(-2)mol//L , what is its concentration after 825 s ? (b) How long would it take for the concentration of N_(2)O_(5) to decrease to 100xx10^(-2) mol L^(-1) from its initiqal value, given in (a) ? Strategy : Since this reaction has a first order rate law, d[N_(2)O_(5)]//dt = k[N_(2)O_(5)] , we can use the corresaponding concentration time equation for a first order reaction : k = (2.303)/(t) log ([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t)) In each part, we substitute the know quantities into this equation and solve for the unkbnown.

The decomposition of N_(2)O_(5) takes place according to I order as: 2N_(2)O_(5) rarr 4NO_(2)+O_(2) Calculate: (a) The rate constant, if instantaneous rate is 1.4 xx 10^(-6) mol litre^(-1) sec^(-1) when concentration of N_(2)O_(5) is 0.04M . (b) The rate of reaction when concentration of N_(2)O_(5) is 1.20 M . (c ) The concentration of N_(2)O_(5) when the rate of reaction will be 2.45 xx 10^(-5) mol litre^(-1) sec^(-1)

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