Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation 2NO+O₂ = 2NO₂2. At high temperatures the dependence of the rate of this reaction on the concentrations of NO, O₂, and NO₂ is complicated. However, at 25°C the rate at which NO₂ is formed obeys the law of mass action and is given by the rate equation dx dt = k(a − x)²(B− 1), where x(1) denotes the concentration of NO₂ at time t, k is the rate constant, a is the initial concentration of NO, and ß is the initial concentration of O₂. At 25°C, the constant k is 7.13 × 10³ (liter)²/(mole)²(second). Let a = 0.0010 mole/L, ß = 0.0041 mole/L, and x(0) = 0 mole/L. Use the fourth-order Runge-Kutta algorithm to approximate x(10). For a tolerance of ε = 0.000001, use a stopping procedure based on the relative error. O x(10) is approximately 2.7 X 10^(-3). O x(10) is approximately 3.5 X 10^(-4). Ox(10) is approximately 3.1 X 10^(-4). O x(10) is approximately 3.9 X 10^(-4). Ox(10) is approximately 2.2 X 10^(-4).
Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation 2NO+O₂ = 2NO₂2. At high temperatures the dependence of the rate of this reaction on the concentrations of NO, O₂, and NO₂ is complicated. However, at 25°C the rate at which NO₂ is formed obeys the law of mass action and is given by the rate equation dx dt = k(a − x)²(B− 1), where x(1) denotes the concentration of NO₂ at time t, k is the rate constant, a is the initial concentration of NO, and ß is the initial concentration of O₂. At 25°C, the constant k is 7.13 × 10³ (liter)²/(mole)²(second). Let a = 0.0010 mole/L, ß = 0.0041 mole/L, and x(0) = 0 mole/L. Use the fourth-order Runge-Kutta algorithm to approximate x(10). For a tolerance of ε = 0.000001, use a stopping procedure based on the relative error. O x(10) is approximately 2.7 X 10^(-3). O x(10) is approximately 3.5 X 10^(-4). Ox(10) is approximately 3.1 X 10^(-4). O x(10) is approximately 3.9 X 10^(-4). Ox(10) is approximately 2.2 X 10^(-4).
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:Chemical Reactions. The reaction between nitrous
oxide and oxygen to form nitrogen dioxide is given by
the balanced chemical equation 2NO+O₂ = 2NO2.
At high temperatures the dependence of the rate of this
reaction on the concentrations of NO, O2, and NO₂ is
complicated. However, at 25°C the rate at which NO₂ is
formed obeys the law of mass action and is given by the
rate equation
dx = x(a - x)² (8-²).
dt
where x(t) denotes the concentration of NO₂ at time
t, k is the rate constant, a is the initial concentration of
NO, and ß is the initial concentration of O₂. At 25°C,
the constant k is 7.13 × 10³ (liter)2/(mole)²(second).
Let a = 0.0010 mole/L, ß = 0.0041 mole/L, and
x(0) = 0 mole/L. Use the fourth-order Runge-Kutta
algorithm to approximate x(10). For a tolerance of
ε = 0.000001, use a stopping procedure based on the
relative error.
x(10) is approximately 2.7 X 10^(-3).
x(10) is approximately 3.5 X 10^(-4).
O x(10) is approximately 3.1 X 10^(-4).
O x(10) is approximately 3.9 X 10^(-4).
O x(10) is approximately 2.2 X 10^(-4).
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