# The following data were obtained for the gas-phase decomposition of dinitrogen pentoxide. 2 N2O5(g) → 4 NO2(g)?

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The following data were obtained for the gas-phase decomposition of dinitrogen pentoxide.

2 N2O5(g) → 4 NO2(g) + O2(g)

[N2O5]0

(mol/L) Initial Rate

(mol/L·s)

0.0750 8.90* 10-4

0.190 2.26* 10-3

0.275 3.26* 10-3

0.410 4.85* 10-3

Defining the rate as −Δ[N2O5]/Δt, How do I determine the rate law?

What is the value of the rate constant?

Please and Thank you. : )

• [N2O5]º(M)…..Initial Rate(M/s)

0.0750…………….8.90×10-4

0.190………………2.26×10-3

0.275………………3.26×10-3

0.410………………4.85×10-3

to determine the rate constant, you first have to determine the rate order. with the balanced equation, it appears that is it second order due to 2 moles N2O5 being present.

when we compare rate2 to rate1, we see that x, the exponent, = 1 so the reaction is first order

rate2 = 2.26×10^-3 = k(0.19)^x

——————————————–

rate1 = 8.9×10^-4 = k(0.075)^x

2.54 = 0.19^x/0.075^x

ln2.54 = x(ln(0.19/0.075))

0.932 = x(0.93)…..x = 1

reaction is first order

rate = k[N2O5]

8.9×10^-4 = k (0.075M)…..k = 0.0119

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