Question

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 1

(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)`