Nitrous oxide N2O undergoes decomposition on a gold surface

Nitrous oxide, N₂O, undergoes decomposition on a gold surface at 900 °C, obeying the following stoichiometry: 2 N₂O _(g) --> 2 N_2(g) +O_{2 (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?  

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What was the initial molarity of the reactant? (N₂O)₀ =

Solution

Decomposition of N₂O in gas phase proceeds in two stages.

Stage 1 : N₂O ----------> N₂ + O

Stage 2 : N₂O + O --------> N₂ + O₂

      Rate of the reaction depends only on concentration of N₂O, hence this is a first order reaction.

Rate = K [A]

Rate = K [N₂O]

ln [A] = ln [A₀] –Kt

According to the graph y = -0.01224 X + -2.902

Graph represents a straight line that intercepts the Y axis at -2.902.

Hence ln [A₀] = -2.902

                 [A₀] = e^(-2.902)

             Initial molarity of the reactant    [A₀] = -7.888 mol/dm³

Numerical value of rate constant , K = 0.01224 (time^-1)