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.

Using the concentration time equation for a first order reaction : The

1.	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.
Answer» (a) Substituting the values of k, t and `[N_(2)O_(5)]_(0)` into the concentration - time equation, we have `4.80xx10^(-4)s^(-1)= (2.303)/(825s)log(1.65xx10^(-2)mol L^(-1))/([N_(2)O_(5)]_(t))` or log `((1.65xx10^(-2)mol L^(-1))/([N_(2)O_(5)]_(t)) = ((4.80xx10^(-4)s^(-1))(825s))/((2.303)))` `= 0.172` or `log (([N_(2)O_(5)]_(t))/(1.65xx10^(-2)mol L^(-1)))= -0.172` To solve for `[N_(2)O_(5)]_(t)` , we take thew antilogarithm of both sides. This removes the long from the left and yield antilog `(-0.172)`, or `10^(-0.172)` , on the right, which equals `0.673`. Thus `([N_(2)O_(5)]_(t))/(1.65xx10^(-2)mol L^(-1)) = 0.673` Hence `[N_(2)O_(5)]_(t) = (1.65xx10^(-2)mol L^(-1)) (0.673)` `= 0.0111 mol L^(-1)` (b) Writing the integrated first order rate equation we have `k = (2.303)/(t) log ([N_(2)O_(5)]_(0)/([N_(2)O_(5)]_(t))) ` or `log (([N_(2)O_(5)]_(t))/([N_(2)O_(5)]_(0))) = (-kt)/(2.303)` Substituting the concentrations and the value of k, we get `log (((1.00xx10^(-2)mol L^(-1)))/((1.65xx10^(-2)mol L^(-1)))) = (-4.80xx10^(-4)s^(-1)xxt)/(2.303)` The left side equals `-0.217` , the right side equals `-2.08xx10^(-4)s^(-1)xxt` . Hence `0.217=2.08xx10^(-4)s^(-1)xxt` or `t = (0.217)/(2.08xx10^(-4)s^(-1))` `= 1.04xx10^(3)`

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