Nitrous oxide, N_2O, undergoes decomposition on a gold surface at 900 degree C,

Question

Nitrous oxide, N_2O, undergoes decomposition on a gold surface at 900 degree C, obeying the following stoichiometry: The concentration of the reactant was monitored as a function of time. A graph of ln(N_20) vs. t gave the graph shown below. The trendline is printed on the box, with y = ln(N_2O) and x = t. The correlation coefficient is r = squareroot R^2 = -0.9995, for a nearly perfect anticorrelation. What is the reaction order according to the above results? ____ What are the units of the rate constant? ____ What is the numerical value of the rate constant? k = ____ What is the half-life for the reaction? t_1/2 = ____ What was the initial molarity of the reactant? (N_2O)_0 = ___

Explanation / Answer

Given

graph is ln(N2O) vs t which gives a straight line with perfect fit hence

it follows First order reaction Answer 1 order

rate law for first order is

rate = k [N2O]

M/s k M

so k will have unit of /s Answer unit of K is s-1

Value of k can be obtained from the straight line equation

y = 1.288 * 10^-2 * x - 2.291

-ln[N2O] = k * t - ln [N2O]intial from first order rate law

k = 1.288 * 10^-2 = 0.01288 Answer k = 0.01288 s-1

Half life for first order reaction

t1/2 = 0.693 / k = 0.693 / 0.01288 = 53.8 s Answer t1/2 = 53.8 s

from the line equation and rate law above

- ln [N2O]intial = -2.291

[N2O] = 9.885 M Answer intial [N2O] = 9.885 M

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