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Chapter 8: Problem 6

The following chemical reactions take place in a closed system \\[\begin{array}{l}2 \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C} \\\\\mathrm{A}+\mathrm{D} \rightleftharpoons \mathrm{C}\end{array}\\] At equilibrium. they can be characterized by \\[K_{1}=\frac{c_{c}}{c_{d}^{2} c_{b}}\\] \\[K_{2}=\frac{c_{c}}{c_{a} c_{d}}\\] where the nomenclature \(c_{i}\) represents the concentration of constituent i. If \(x_{1}\) and \(x_{2}\) are the number of moles of \(C\) that are produced due to the first and second reactions, respectively, use an approach similar to that of Prob. 8.5 to reformulate the equilibrium relationships in terms of the initial concentrations of the constituents. Then, use the Newton-Raphson method to solve the pair of simultaneous nonlinear equations for \(x_{1}\) and \(x_{2}\) if \(K_{1}=4 \times 10^{-4}, K_{2}=3.7 \times 10^{-2}\) \(c_{u, 0}=50, c_{b, 0}=20, c_{c .0}=5,\) and \(c_{d, 0}=10 .\) Use a graphical approach to develop your initial guesses.

Short Answer

Expert verified
In this problem, we have two chemical reactions occurring simultaneously in a closed system. We express the equilibrium relationships in terms of the initial concentrations and substitute these into the equilibrium constant expressions. Then, we use a graphical approach to determine initial guesses for the Newton-Raphson method. Finally, we iteratively solve the simultaneous nonlinear equations for \(x_1\) and \(x_2\), which represent the number of moles of C produced by each reaction, using the Newton-Raphson method until our estimates converge to the desired level of accuracy.

Step by step solution

01

Express the equilibrium relationships in terms of initial concentrations

For reaction 1: \(2A + B \rightleftharpoons C\) Let \(x_1\) moles of C be produced. At equilibrium, we have these concentrations: \ \[c_A = c_{u0} - 2 x_1\] \[c_B = c_{b0} - x_1\] \[c_C = c_{c0} + x_1\] For reaction 2: \(A + D \rightleftharpoons C\) Let \(x_2\) moles of C be produced. At equilibrium, we have these concentrations: \ \(c_A = c_{u0} - 2 x_1 - x_2\) \[c_D = c_{d0} - x_2\] \[c_C = c_{c0} + x_1 + x_2\] Now, we substitute these concentrations into the equilibrium constant expressions.
02

Replace concentrations in the equilibrium constant expressions

For \(K_1\): \[K_1 = \frac{c_c}{c_a^2 c_b} = \frac{c_{c0} + x_1}{(c_{u0} - 2x_1)^2 (c_{b0} - x_1)}\] For \(K_2\): \[K_2 = \frac{c_c}{c_a c_d} = \frac{c_{c0} + x_1 + x_2}{(c_{u0} - 2x_1 - x_2)(c_{d0} - x_2)}\] Now we have two equations in terms of \(x_1\) and \(x_2\), and we need to solve these simultaneous nonlinear equations.
03

Use graphical approach to develop initial guesses

In order to develop initial guesses for the Newton-Raphson method, we can plot the two functions for a range of values of \(x_1\) and \(x_2\). By observing where the graphs intersect, we can have a reasonable initial guess for the iterative method.
04

Solve the simultaneous nonlinear equations using the Newton-Raphson method

Now that we have our initial guesses, we can use the Newton-Raphson method to solve the simultaneous nonlinear equations for \(x_1\) and \(x_2\). We will need to develop the Jacobian matrix, which consists of the partial derivatives of the two equations with respect to \(x_1\) and \(x_2\). Once we have that, we will iteratively update our estimates for \(x_1\) and \(x_2\) until the equations converge to the desired level of accuracy. The final solution will be the values of \(x_1\) and \(x_2\) that satisfactorily satisfy the simultaneous nonlinear equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton-Raphson Method
The Newton-Raphson method is a powerful technique commonly used to find approximate solutions to nonlinear equations. It is an iterative process that starts from an initial guess which is then improved repeatedly until a sufficiently accurate solution is found.

For the given problem, this method is particularly useful because we're dealing with simultaneous nonlinear equations representing chemical reaction equilibria. The strategy involves using the derivative (or Jacobian matrix for systems of equations) to predict the behavior of the functions and rapidly converge to the roots. Here's a simplified breakdown of the process:
  • Choose an initial guess for the values of the variables, in this case, the number of moles of product C, denoted as \(x_1\) and \(x_2\).
  • Calculate the value of the function (or functions, for systems) at this guess.
  • Also calculate the derivative (or Jacobian matrix) at this point.
  • Use these calculations to determine the next best estimate for the variable(s).
  • Repeat the process until the changes in the guesses are smaller than a pre-defined threshold, indicating that you've likely found a solution near the actual root.
Each iteration brings the estimate closer to the actual solution. The method assumes that the functions are differentiable and that the initial guess is close enough to the actual root that the method converges.

Nonlinear Equations
Nonlinear equations are equations of a variable where the variable is not simply raised to the first power. Such equations do not form straight lines when graphed and often feature more complex behaviors such as curves, bends, and asymptotes.

In chemical processes, these equations can describe how reactants transform into products and help us understand reaction rates, equilibrium, and other vital information. The challenge with nonlinear equations lies in the fact that they do not have a straightforward, analytic solving method, like simple algebraic equations do. Thus, numerical methods like the Newton-Raphson technique are particularly handy.

For the exercise dealing with chemical reactions, the concentrations of reactants and products change as the reaction proceeds, resulting in the formation of nonlinear equations that relate these concentrations to each other and to the equilibrium constants \(K_1\) and \(K_2\). To solve these equations for the variables \(x_1\) and \(x_2\), which signify the moles of product formed, one must employ an iterative technique as direct solving is impractical.

Chemical Reaction Equilibrium
Chemical reaction equilibrium is a state in a chemical process where the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of reactants and products over time. This is represented mathematically by the equilibrium constant expression. The larger the equilibrium constant, the more the reaction favors product formation.

In the exercise, the equilibrium constants \(K_1\) and \(K_2\) describe two separate reactions happening within the same system and offer a quantifiable measure of how far the reactions proceed. These constants are useful in calculating unknown concentrations once the system has reached equilibrium.

The reformulation of the equilibrium expressions in terms of the initial concentrations allows for a more practical approach to solving for the variables \(x_1\) and \(x_2\), representing the production of substance C. Once the nonlinear equations are formulated in this manner, predicting how changes in the initial concentrations of reactants affect the equilibrium becomes possible, making it a powerful tool for chemists and engineers.

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Most popular questions from this chapter

The pH of water has great significance to environmental and chemical engineers. It can be related to processes ranging from pipe corrosion to*acid rain. The pH is related to the hydrogen ion concentration by \\[\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]\\] The following five equations govern the concentrations of \(a\) mixture of carbon dioxide and water for a closed system: \\[\begin{array}{l}K_{1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]} \\\ K_{2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]}{\left[\mathrm{HCO}_{3}^{-}\right]} \\\ K_{w}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ c_{T}=\left[\mathrm{CO}_{2}\right]+\left[\mathrm{HCO}_{3}^{-}\right]+\left[\mathrm{CO}_{3}^{2-}\right] \\\ \mathrm{Alk}=\left[\mathrm{HCO}_{3}^{-}\right]+2\left[\mathrm{CO}_{3}^{2-}\right]+\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]\end{array}\\] where \(A I k=\) the alkalinity, \(c_{T}=\) total inorganic carbon, and the \(K^{\prime}\) s are equilibrium coefficients. The five unknowns are \(\left[\mathrm{CO}_{2}\right]=\) carbon dioxide, \(\left[\mathrm{HCO}_{3}^{-}\right]=\) bicarbonate, \(\left[\mathrm{CO}_{3}^{2-}\right]=\) carbonate \(\left[\mathrm{H}^{+}\right]=\) hydrogen ion, and \(\left[\mathrm{OH}^{-}\right]=\) hydroxyl ion. Solve for the five unknowns given that Alk \(=2 \times 10^{-3}, c r=3 \times 10^{-3}\) \(K_{1}=10^{-6.3}, K_{2}=10^{-103},\) and \(K_{w}=10^{-14} .\) Also, calculate the solution's pH.

The general form for a three-dimensional stress field is given by \\[\left[\begin{array}{lll}\sigma_{x x} & \sigma_{x y} & \sigma_{x z} \\ \sigma_{x y} & \sigma_{y y} & \sigma_{y z} \\\\\sigma_{x z} & \sigma_{y z} & \sigma_{z z}\end{array}\right]\\] where the diagonal terms represent tensile or compressive stresses and the off-diagonal terms represent shear stresses. A stress field (in MPa) is given by \\[\left[\begin{array}{ccc}10 & 14 & 25 \\\14 & 7 & 15 \\\25 & 15 & 16\end{array}\right]\\] To solve for the principal stresses, it is necessary to construct the following matrix (again in MPa): \\[\left[\begin{array}{ccc}10-\sigma & 14 & 25 \\\14 & 7-\sigma & 15 \\\25 & 15 & 16-\sigma\end{array}\right]\\] \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) can be solved from the equation \\[\sigma^{3}-I \sigma^{2}+I I \sigma-I I I=0\\] where \\[\begin{array}{l}I=\sigma_{x x}+\sigma_{y y}+\sigma_{z z} \\\I I=\sigma_{x x} \sigma_{y y}+\sigma_{x x} \sigma_{z z}+\sigma_{y y} \sigma_{zz}-\sigma_{x y}^{2}-\sigma_{x z}^{2}-\sigma_{y z}^{2} \\\I I=\sigma_{x x} \sigma_{y y} \sigma_{z z}-\sigma_{x x} \sigma_{y z}^{2}-\sigma_{y y} \sigma_{x z}^{2}-\sigma_{z z} \sigma_{x y}^{2}+2 \sigma_{x y} \sigma_{x z} \sigma_{y z}\end{array}\\] \(I, I I,\) and \(I I I\) are known as the stress invariants. Find \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) using a root-finding technique.

In Sec. \(8.4,\) the phase angle \(\phi\) between the forced vibration caused by the rough road and the motion of the car is given by \\[\tan \phi=\frac{2\left(c / c_{c}\right)(\omega / p)}{1-(\omega / p)^{2}}\\] As a mechanical engineer, you would like to know if there are cases where \(\phi=\omega / 3-1 .\) Use the other parameters from the section to set up the equation as a roots problem and solve for \(\omega\).

Real mechanical systems may involve the deflection of nonlinear springs. In Fig. \(\mathrm{P} 8.34\), a mass \(m\) is released a distance \(h\) above a nonlinear spring. The resistance force \(F\) of the spring is given by \\[F=-\left(k_{1} d+k_{2} d^{3 / 2}\right)\\] Conservation of energy can be used to show that \\[0=\frac{2 k_{2} d^{5 / 2}}{5}+\frac{1}{2} k_{1} d^{2}-m g d-m g h\\] Solve for \(d\), given the following parameter values: \(k_{1}=50,000 \mathrm{g} / \mathrm{s}^{2}\) \\[k_{2}=40 g /\left(s^{2} \mathrm{m}^{0.5}\right), m=90 \mathrm{g}, g=9.81 \mathrm{m} / \mathrm{s}^{2}, \text { and } h=0.45\mathrm{m}\\]

The space shuttle, at lift-off from the launch pad, has four forces acting on it, which are shown on the free-body diagram (Fig. P8.46). The combined weight of the two solid rocket boosters and external fuel tank is \(W_{B}=1.663 \times 10^{6} \mathrm{lb}\). The weight of the orbiter with a full payload is \(W_{s}=0.23 \times 10^{6} \mathrm{Jb}\). The combined thrust of the two solid roctet boosters is \(T_{B}=5.30 \times 10^{6}\) lb. The combined thrust of the three liquid fuel orbiter engines is \(T_{S}=1.125 \times 10^{6} \mathrm{lb}\) At liftoff, the orbiter engine thrust is directed at angle \(\theta\) to make the resultant moment acting on the entire craft assembly (external tank, solid rocket boosters, and orbiter) equal to zero. With the resultant moment equal to zero, the craft will not rotate about its mass center \(G\) at liftoff. With these forces, the craft will have a resultant force with components in both the vertical and horizontal direction. The vertical resultant force component is what allows the craft to lift off from the launch pad and fly vertically. The horizontal resultant force component causes the craft to fly horizontally. The resultant mornent acting an the craft will be zero when \(\theta\) is adjusted to the proper value. If this angle is not adjusted properly, and there is some resultant moment acting on the craft, the craft will tend to rotate about it mass center. (a) Resolve the orbiter thrust \(T_{S}\) into horizontal and vertical components, and then sum moments about point \(G,\) the craft mass center. Set the resulting moment equation equal to zero. This equation can now be solived for the value of \(\theta\) required for liftoff. (b) Derive an equation for the resutent moment acting on the craft in terms of the angle \(\theta\). Plot the resultant moment as a function of the angle \(\theta\) over a range of -5 radians to +5 radians. (c) Write a computer program to solve for the angle \(\theta\) using Newton's method to find the root of the resultant moment equation. Make an initial first guess at the root of interest using the plot. Terminate your iterations when the value of \(\theta\) has better than five significant figures. (d) Repeat the program for the minimum payload weight of the orbiter of \(W_{S}=195,000 \mathrm{lb}\).
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