The decomposition of in solution in carbon tetrachloride proceeds via the reaction 2 N₂O₅(soln) → 4 NO₂(soln) + O₂(soln)
The reaction is first order and has a rate constant of 4.82 × 10^-3s^-1 at 64°C.If the reaction is initiated with 0.058 mol in a 1.00-L vessel,how many moles remain after 151 s?

Question

The decomposition of

in solution in carbon tetrachloride proceeds via the reaction 2 N₂O₅(soln) → 4 NO₂(soln) + O₂(soln)
The reaction is first order and has a rate constant of 4.82 × 10^-3s^-1 at 64°C.If the reaction is initiated with 0.058 mol in a 1.00-L vessel,how many moles remain after 151 s?

Quizplus · Accepted Answer

Using first-order rate law:

ln([A]t/[Ao])=-kt

Where [A]t is the concentration at time t, [Ao] is the initial concentration, k is the rate constant, and t is time.

Rearranging the equation we get:

[A]t=[Ao]e^(-kt)

Plugging in the values given in the problem, we get:

[A]t=(0.058 mol)(e^(-4.82x10^-3 mol/L*s x 151 s)) 
[A]t=0.028 mol/L or 0.028 M

Therefore, the answer is C, 0.028 M.

The Decomposition of in Solution in Carbon Tetrachloride