Question

In: Chemistry

***Given the following*** Balanced Equation: 2 N2O5 -> 4 NO2 + O2 Rate Constant, k= 4.66...

***Given the following***

Balanced Equation: 2 N2O5 -> 4 NO2 + O2

Rate Constant, k= 4.66 x 10-4 s-1

Temperature: 72Degrees Celcius

***How much N2O5 remains in the solution after 115 minutes if the initial concentration of N2O5 is 0.60 M?

***What % of N2O5 has reacted at this point?

If you're able to show your work for me to get a better understanding it would be greatly appreciated! Thank you!

Solutions

Expert Solution

2N2O5 ------> 4NO2 + O2

rate constant unit shows it is first order reaction

rate = k [ N2O5 ]

integrated form of first order rate equation is

ln[A]t= -kt + ln[A]0

where,

  [A]t= concentration of reactant at time t

t = time , 115minutes = 6900seconds

[A]0= Initial concentration, 0.60M

k = rate constant , 4.66×10^-4s^-1

Therefore,

ln[A]t = -(4.66×10^-4s^-1 × 6.9×10^3s) + ln(0.60M)

2.303log[A]t= -(4.66×10^-4s^-1×6.9×10^3s)+ 2.303log(0.60M)

= -3.2154 - 0.5109

= - 3.7263

log[A]t= -1.6180

[A]t = 0.0241M

So, after 115minutes , 0.0241M of N2O5 will remains

Initial concentration of N2O5 = 0.60M

N2O5 remains after 115 minutes = 0.0241M

N2O5 reacted = 0.60M - 0.0241M = 0.5759M

℅ of N2O5 reacted = (0.5759M/0.60M)×100 = 95.98%

queen_honey_blossom answered 2 years ago
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