Question

Chemical Reactions.The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation\({\bf{2NO + }}{{\bf{O}}_{\bf{2}}}{\bf{ = 2N}}{{\bf{O}}_{\bf{2}}}\). At high temperatures the dependence of the rate of this reaction on the concentrations of NO,\({{\bf{O}}_{\bf{2}}}\), and \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) is complicated. However, at \({\bf{2}}{{\bf{5}}^{\bf{o}}}\)C the rate at which NO2 is formed obeys the law of mass action and is given by the rate equation\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = k(\alpha - x}}{{\bf{)}}^{\bf{2}}}\left( {{\bf{\beta - }}\frac{{\bf{x}}}{{\bf{2}}}} \right)\), where \({{\bf{x}}_{\bf{1}}}{{\bf{t}}_{\bf{2}}}\) denotes the concentration of \({\bf{N}}{{\bf{o}}_{\bf{2}}}\) at time t, kis the rate constant, a is the initial concentration of NO, and b is the initial concentration of \({{\bf{O}}_{\bf{2}}}\). At \({\bf{2}}{{\bf{5}}^{\bf{o}}}\)C, the constant kis \({\bf{7}}{\bf{.13 \times 1}}{{\bf{0}}^{\bf{3}}}{{\bf{(litre)}}^{\bf{2}}}{\bf{/(mole}}{{\bf{)}}^{\bf{2}}}\) (second).Let \({\bf{\alpha = 0}}{\bf{.0010}}\) mole/L, \({\bf{\beta = 0}}{\bf{.0041}}\) mole/L, and (\({\bf{x(0) = 0}}\) mole>L. Use the fourth-order Runge–Kutta algorithm to approximate \({\bf{x}}\left( {{\bf{10}}} \right)\). For a tolerance of\({\bf{\varepsilon = 0}}{\bf{.000001}}\), use a stopping procedure based on the relative error.

Short answer

The approximation value is \(\phi \left( {10} \right) = 0.000223596\).

Step-by-step solution

Important hint.

To get the solution use the concept of Runge-Kutta algorithm.

Find the values for \({\bf{t}}\) and \({\bf{x}}\).

Using the improved 4^th order Runge-Kutta subroutine with tolerance \(\varepsilon = 0.000001\).

Since \(f\left( {t,x} \right) = 7.13 \times {10^3}{\left( {0.0010 - x} \right)^2}\left( {0.0041 - \frac{x}{2}} \right)\)

and \(t = {t_0} = 0,x = {x_o} = 0\) and\(h = 10\)

\(\begin{array}{c}{k_1} = hf\left( {t,x} \right)\\ = h\left[ {7.13 \times {{10}^3}{{\left( {0.0010 - x} \right)}^2}\left( {0.0041 - \frac{x}{2}} \right)} \right]\end{array}\)

\(\begin{array}{c}{k_2} = hf\left( {t + \frac{h}{2},x + \frac{{{k_1}}}{2}} \right)\\ = h\left[ {7.13 \times {{10}^3}{{\left( {0.0010 - x - \frac{{{k_1}}}{1}} \right)}^2}\left( {0.0041 - \frac{{x + \frac{{{k_1}}}{2}}}{2}} \right)} \right]\end{array}\)

\(\begin{array}{c}{k_3} = hf\left( {t + \frac{h}{2},x + \frac{{{k_2}}}{2}} \right)\\ = h\left[ {7.13 \times {{10}^3}{{\left( {0.0010 - x - \frac{{{k_2}}}{1}} \right)}^2}\left( {0.0041 - \frac{{x + \frac{{{k_2}}}{2}}}{2}} \right)} \right]\end{array}\)

\(\begin{array}{c}{k_4} = hf\left( {t + h,x + {k_3}} \right)\\ = h\left[ {7.13 \times {{10}^3}{{(0.0010 - x - h)}^2}\left( {0.0041 - \frac{{x + {k_3}}}{2}} \right)} \right]\end{array}\)

\(\begin{array}{c}{k_1} = h\left( {t,x} \right)\\ = 0.00029233\end{array}\)

\(\begin{array}{c}{k_2} = hf\left( {t + \frac{h}{2},x + \frac{{{k_1}}}{2}} \right)\\ = 0.00020932\end{array}\)

\(\begin{array}{c}{k_3} = hf\left( {t + \frac{h}{2},x + \frac{{{k_2}}}{2}} \right)\\ = 0.000231351\end{array}\)

\(\begin{array}{c}{k_4} = hf\left( {t + h,x + {k_3}} \right)\\ = 0.000167842\end{array}\)

Find the values for \({\bf{t}}\) and \({\bf{x}}\).

\(\begin{array}{c}t = {t_o} + h\\t = 10\\x = {x_o} + \frac{1}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)\\ = 0.000223585\end{array}\)

Therefore, \(\phi \left( {10} \right) = 0.000223585\).

This the solution of IVP.

Now, the relative error is

\(\begin{array}{c}\varepsilon = \left| {\frac{{0.0002235885 - 0}}{{0.0002235885}}} \right|\\ = 1 > 0.000001\end{array}\)

Find the other values.

Apply the same procedure for \(h = 5\),

\(t = 0,x = 0\).

\(\begin{array}{c}{k_1} = h\left( {t,x} \right)\\ = 0.000146165\end{array}\)

\(\begin{array}{c}{k_2} = hf\left( {t + \frac{h}{2},x + \frac{{{k_1}}}{2}} \right)\\ = 0.000124462\end{array}\)

\(\begin{array}{c}{k_3} = hf\left( {t + \frac{h}{2},x + \frac{{{k_2}}}{2}} \right)\\ = 0.000127564\end{array}\)

\(\begin{array}{c}{k_4} = hf\left( {t + h,x + {k_3}} \right)\\ = 0.000109522\end{array}\)

\(\begin{array}{c}t = 0 + 5 = 5\\x = 0 + \frac{1}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)\\ = 0.000126623\end{array}\)

\(\begin{array}{c}{k_1} = h\left( {t,x} \right)\\ = 0.000109771\end{array}\)

\(\begin{array}{c}{k_2} = hf\left( {t + \frac{h}{2},x + \frac{{{k_1}}}{2}} \right)\\ = 0.0000957526\end{array}\)

\(\begin{array}{c}{k_3} = hf\left( {t + \frac{h}{2},x + \frac{{{k_2}}}{2}} \right)\\ = 0.0000974847\end{array}\)

\(\begin{array}{c}{k_4} = hf\left( {t + h,x + {k_3}} \right)\\ = 0.0000855878\end{array}\)

\(\begin{array}{c}t = 5 + 5 = 10\\x = 0 + \frac{1}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)\\ = 0.000223595\end{array}\)

Thus the values is, \(\phi \left( {10} \right) = 0.000223595\)[ss1] [m2]

The relative error is

\(\begin{array}{c}\varepsilon = \left| {\frac{{0.000223595 - 0.000223585}}{{0.000223595}}} \right|\\ = 0.0000447237 > 0.0000001\end{array}\)

[ss1]Incomplete sentance

[m2]done

Evaluate the other values.

Apply the same procedure for other values. The values are

\(\begin{array}{c}\phi \left( {10} \right) = v\left( {10;2.5} \right)\\ = 0.000223596\\\varepsilon = \left| {\frac{{0.000223596 - 0.00023595}}{{0.000223596}}} \right|\\ = 0.00000447235 > 0.000001\\\phi \left( {10} \right) = v\left( {10;1.25} \right)\\ = 0.000223596\\\varepsilon = \left| {\frac{{0.000223596 - 0.000223596}}{{0.000223596}}} \right|\\ = 0 < 0.0001\end{array}\)

Hence, the value is \(\phi \left( {10} \right) = 0.000223596\).[ss1] [m2]

[ss1]Incomplete sentence

[m2]done