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This problem deals with a chemical reaction following the law of mass action, where the rate of reaction depends on the concentrations of two reactants \( A \) and \( B \), which form a product \( C \). The given rate equation is:

\[
\frac{d[C]}{dt} = k[A][B]
\]

This is rewritten in terms of \( x = [C] \), the concentration of the product \( C \), and initial concentrations of \( A \) and \( B \) as \( a \) and \( b \), respectively:

\[
\frac{dx}{dt} = k(a - x)(b - x)
\]

### Part (a)
We need to find \( x(t) \) assuming \( a \neq b \), with the initial condition \( x(0) = 0 \). To solve this differential equation:

1. Separate the variables:
   \[
   \frac{1}{(a - x)(b - x)} dx = k \, dt
   \]

2. Perform partial fraction decomposition to split the left-hand side:
   \[
   \frac{1}{(a - x)(b - x)} = \frac{1}{a - b} \left( \frac{1}{a - x} - \frac{1}{b - x} \right)
   \]

3. Integrate both sides:
   \[
   \frac{1}{a - b} \left( \ln \left| \frac{a - x}{b - x} \right| \right) = kt + C
   \]

   Since \( x(0) = 0 \), use the initial condition to solve for \( C \):
   \[
   \frac{1}{a - b} \left( \ln \left| \frac{a}{b} \right| \right) = C
   \]
   So,
   \[
   C = \frac{1}{a - b} \ln \left( \frac{a}{b} \right)
   \]

4. After simplifying, the expression for \( x(t) \) becomes:
   \[
   \ln \left| \frac{(a - x)(b)}{(b - x)(a)} \right| = k(a - b)t
   \]
   Finally, solve for \( x(t) \).

### Part (b)
In this case, \( a = b \). The differential equation simplifies to:

\[
\frac{dx}{dt} = k(a - x)^2
\]

This can be solved by separating variables:

\[
\frac{1}{(a - x)^2} dx = k \, dt
\]

Integrating both sides:

\[
\int \frac{1}{(a - x)^2} dx = \int k \, dt
\]

The left side integrates to:
\[
\frac{-1}{a - x} = kt + C
\]

Using the initial condition \( x(0) = 0 \), we find \( C \). The solution is then:

\[
x(t) = a \left( 1 - \frac{1}{1 + ak t} \right)
\]

Finally, we simplify for \( x(t) \) given that after 20 seconds, \( x(20) = \frac{a}{2} \), and solve for \( k \).

Would you like further explanation or details on any step? 

Here are 5 related questions to expand on this topic:
1. How does the solution change if we assume the reaction is reversible?
2. How can we apply this type of rate equation to more complex reactions with multiple steps?
3. What would happen to the rate equation if a catalyst were introduced?
4. How do the initial concentrations of \( A \) and \( B \) affect the time to reach equilibrium?
5. Can you solve the equation for \( x(t) \) numerically using given values of \( k \), \( a \), and \( b \)?

**Tip:** Understanding how to manipulate differential equations using separation of variables is crucial for solving many reaction-rate problems in chemistry.

In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B -> C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B. Assuming [A] = a moles/L and [B] = b moles/L, find x(t), the concentration of the product C, for both cases where a ≠ b and a = b.

Learn how to solve a chemical reaction rate equation using differential equations. This problem involves finding the concentration of product C over time, considering different initial reactant concentrations of A and B. It covers key mathematical concepts such as separation of variables and partial fraction decomposition.

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