Question

1A. The decomposition of dinitrogen pentoxide in carbon tetrachloride solution at 30 °C N2O5 2 NO2 + ½ O2 is first order in N2O5 with a rate constant of 4.10×10-3 min-1. If the initial concentration of N2O5 is 0.510 M, the concentration of N2O5 will be M after 402 min h

1B. The gas phase decomposition of phosphine at 120 °C

PH₃(g)1/4 P₄(g) + 3/2 H₂(g)

is first order in PH₃ with a rate constant of 1.80×10^-2 s^-1.

If the initial concentration of PH₃ is 5.08×10^-2 M, the concentration of PH₃ will be 1.06×10^-2 M after s have passed.

1C. The decomposition of nitrous oxide at 565 °C

N₂O(g)N₂(g) + ½ O₂(g)

is second order in N₂O.

In one experiment, when the initial concentration of N₂O was 1.10 M, the concentration of N₂O dropped to 0.151 M after 3.42×10³ seconds had passed.

Based on these data, the rate constant for the reaction is M^-1 s^-1.

1D. The gas phase decomposition of nitrogen dioxide at 383 °C

NO₂(g)NO(g) + ½ O₂(g)

is second order in NO₂ with a rate constant of 0.540 M^-1 s^-1.

If the initial concentration of NO₂ is 5.12×10^-2 M, the concentration of NO₂ will be M after 230 seconds have passed.

Answer 1

Q1A. concentration of N₂O₅ = 0.0981 M

Q1B. time = 87.1 s

Q1C. rate constant = 1.67 x 10^-3 M^-1 s^-1

Q1D. concentration of NO₂ = 6.96 x 10^-3M

Explanation

Q1A. The integrated first order rate law is given as :

ln[A] = ln[A]_o - (k) * (t)

where [A]_o = initial reactant concentration = 0.510 M

[A] = final reactant concentration

k = rate constant = 4.10 x 10^-3 min^-1

t = time passed = 402 min

Substituting the values,

ln[A] = ln(0.510 M) - (4.10 x 10^-3 min^-1) * (402 min)

ln[A] = -2.32

[A] = e^-2.32

[A] = 0.0981 M