Nitrous oxide, N 2 O, undergoes decomposition on a gold surface at 900 °C, obeyi

Question

Nitrous oxide, N2O, undergoes decomposition on a gold surface at 900 °C, obeying the following stoichiometry: 2 N2O (g) --> 2 N2(g) +O2 (g)

The concentration of the reactant was monitored as a function of time. A graph of ln(N2O) vs. t gave a graph simlilar to the one in the lab manual. The trendline is Y= -0.01224 X + -2.902, with a y axis of ln(N2O) and x of time. The correlation coefficient r is -0.9996, for a nearly perfect anticorrelation.

What is the numerical value of the rate constant, k?

What is the reaction order according to the above results?

Select one:

A.

First order

B. Unable to determine from the given data .

C.

Zero order

D.

Second order

Explanation / Answer

ln[A] = -kt + ln[A0]

the graph follows the above equation

a) first order

b) units of k =min-1

c) slope = -k = -1.288E-02

k=1.288E-02

d) t(1/2)=0.693/k = 53.80 min

e) intercept = ln [Ao] = -2.291

[Ao] = 0.1011 M

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