Chapter 10: Problem 35
The following chemical reactions take place in a liquid-phase batch reactor of constant volume \(V\). $$\begin{aligned} &\mathrm{A} \rightarrow 2 \mathrm{B} \quad r_{1}[\mathrm{mol} \mathrm{A} \text { consumed } /(\mathrm{L} \cdot \mathrm{s})]=0.100 C_{\mathrm{A}}\\\ &\mathbf{B} \rightarrow \mathbf{C} \quad r_{2}[\mathrm{mol} \mathbf{C} \text { generated } /(\mathbf{L} \cdot \mathbf{s})]=0.200 C_{\mathrm{B}}^{2} \end{aligned}$$ where the concentrations \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) are in mol/L. The reactor is initially charged with pure \(\mathrm{A}\) at a concentration of 1.00 mol/L. (a) Write expressions for ( \(i\) ) the rate of generation of \(\mathrm{B}\) in the first reaction and (ii) the rate of consumption of \(\mathrm{B}\) in the second reaction. (If this takes you more than about 10 seconds, you're missing the point.) (b) Write mole balances on A, B, and C, convert them into expressions for \(d C_{\mathrm{A}} / d t, d C_{\mathrm{B}} / d t\), and \(d C_{\mathrm{C}} / d t,\) and provide boundary conditions. (c) Without doing any calculations, sketch on a single graph the plots you would expect to obtain of \(C_{\mathrm{A}}\) versus \(t, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t .\) Clearly show the function values at \(t=0\) and \(t \rightarrow \infty\) and the curvature (concave up, concave down, or linear) in the vicinity of \(t=0 .\) Briefly explain your reasoning. (d) Solve the equations derived in Part (b) using a differential equation- solving program. On a single graph, show plots of \(C_{\mathrm{A} \text { versust }}, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t\) from \(t=0\) to \(t=50\) s. Verify that your predictions in Part (c) were correct. If they were not, change them and revise your explanation.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Batch Reactor
This means that the concentrations of reactants and products change over time as the reaction progresses. Since the volume remains constant, it simplifies the calculations and analysis.
- No inflow or outflow: Everything happens in a closed vessel.
- Constant volume: Easy to calculate concentration changes.
- Time-dependent reactions: Ideal for detailed kinetic studies.
Mole Balance
Mole balances are applied to every species in the reaction:
- Conservation of mass: Total mass is constant, only distribution changes.
- Mole balance equations: Derived from reaction kinetics.
- Account for generation and consumption: Ensures accurate modeling.
Rate of Reaction
For example, in the given reactions:
- From A to B: The rate of generation of B, \( r_1 \), is directly proportional to the concentration of A, given by \( 0.100 C_A \).
- From B to C: The rate of consumption of B, \( r_2 \), depends on the square of B's concentration, \( 0.200 C_B^2 \).
Differential Equations
The equations derived for concentration changes are:
- For A: \( \frac{dC_A}{dt} = -0.100 C_A \)
- For B: \( \frac{dC_B}{dt} = 0.200 C_A - 0.200 C_B^2 \)
- For C: \( \frac{dC_C}{dt} = 0.200 C_B^2 \)