Question

Chemical Reactions. A second order chemical reaction involves the interaction (collision) of one molecule of a substance \(P\) with one molecule of a substance \(Q\) to produce one molecule of a new substance \(X ;\) this is denoted by \(P+Q \rightarrow X\). Suppose that \(p\) and \(q\), where \(p \neq q,\) are the initial concentrations of \(P\) and \(Q,\) respectively, and let \(x(t)\) be the concentration of \(X\) at time \(t\). Then \(p-x(t)\) and \(q-x(t)\) are the concentrations of \(P\) and \(Q\) at time \(t,\) and the rate at which the reaction occurs is given by the equation $$ d x / d t=\alpha(p-x)(q-x) $$ where \(\alpha\) is a positive constant. (a) If \(x(0)=0\), determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and find \(x(t)\) for any \(l .\) (b) If the substances \(P\) and \(Q\) are the same, then \(p=q\) and \(\mathrm{Eq}\). (i) is replaced by $$ d x / d t=\alpha(p-x)^{2} $$ If \(x(0)=0,\) determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and determine \(x(t)\) for any \(t .\)

Short answer

In summary, for a second-order chemical reaction between substances P and Q producing substance X, the concentration of substance X as a function of time when p ≠ q is: $$x(t) = \frac{pq(1 - e^{-(p-q)\alpha t})}{q + p(e^{-(p-q)\alpha t} - e^{-(p-q)\alpha t})}$$ with the limiting concentration as \(t \rightarrow \infty\) given by \(\min(p,q)\). When substances P and Q are the same (p = q), the concentration of substance X as a function of time is: $$x(t) = p\left(1 - \frac{1}{1 + \alpha pt}\right)$$ with the limiting concentration as \(t \rightarrow \infty\) given by \(p\).

Step-by-step solution

Find the limiting value of x(t) as t → ∞

We know that as the reaction progresses, the concentrations of \(P\) and \(Q\) are decreasing, so \((p - x(t))\) and \((q - x(t))\) are both decreasing. Consequently, \((p - x(t))(q - x(t))\) (the rate) also decreases as the reaction continues. Since the rate is decreasing, \(dx/dt\) will eventually approach zero. This means that as the reaction progresses, the concentration of \(X\) stops changing, and we reach a limiting value for \(x(t)\). Therefore, we can set \(dx/dt = 0\) and solve for \(x\): $$0 = \alpha(p - x)(q - x)$$ To find the limiting value of \(x(t)\), we can solve for \(x\): $$(p - x)(q - x) = 0$$ Since we are given \(p \neq q\) and \(x(0) = 0\), the limiting value of \(x(t)\) must be the smaller of \(p\) and \(q\) (otherwise it would contradict the initial condition). Therefore, $$x_{lim} = \min(p,q)$$

Solve the initial value problem

Now, we will solve the initial value problem: $$\frac{dx}{dt} = \alpha(p - x)(q - x), \quad x(0) = 0$$ We can rewrite this differential equation as: $$\frac{dx}{(p - x)(q - x)} = \alpha dt$$ To solve this equation, we can integrate both sides: $$\int_{0}^{x}\frac{dy}{(p - y)(q - y)} = \alpha \int_{0}^{t}ds$$ Now, we need to solve the integral on the left side of the equation. We do this using partial fractions: $$\frac{A}{(p - y)} + \frac{B}{(q - y)} = \frac{1}{(p - y)(q - y)}$$ To solve for \(A\) and \(B\), we first multiply both sides by \((p - y)(q - y)\) and then substitute \(y = p\) and \(y = q\) and solve the resulting system of linear equations: $$A(q - p) = 1$$ $$B(p - q) = 1$$ $$A = \frac{1}{(q - p)}, \quad B = \frac{1}{(p - q)}$$ Now, plugging the values of \(A\) and \(B\) back into the integral's equation, we get: $$\int_{0}^{x}\frac{1}{(q - p)(p - y)} - \frac{1}{(p - q)(q - y)} dy = \alpha \int_{0}^{t}ds$$ Integrating, we get: $$\left[-\frac{1}{(q - p)} \ln|p - y| + \frac{1}{(p - q)} \ln|q - y| \right]_0^x = \alpha t$$ Now, we substitute in the bounds of the integral and solve for \(x(t)\): $$\left[-\frac{1}{(q - p)} \ln|p - x| + \frac{1}{(p - q)} \ln|q - x| + \frac{1}{(p - q)} \ln p - \frac{1}{(p - q)} \ln q\right] = \alpha t$$ Next, we solve for \(x(t)\): $$\frac{1}{p-q}\left[\ln\left(\frac{(q-x)p}{(p-x)q}\right)\right]=\alpha t$$ $$\frac{(q-x)p}{(p-x)q} = e^{(p-q)\alpha t}$$ $$\frac{q-x}{p-x}= e^{-(p-q)\alpha t}$$ $$(q-x)=(p-x)e^{-(p-q)\alpha t}$$ $$x(t) = \frac{pq(1 - e^{-(p-q)\alpha t})}{q + p(e^{-(p-q)\alpha t} - e^{-(p-q)\alpha t})}$$ Now we have the concentration of substance \(X\) at any time, \(x(t)\), and the limiting concentration, which is \(\min(p,q)\). ###Part (b):###

Find the limiting value of x(t) as t → ∞ for p = q

Now we have the case where substances \(P\) and \(Q\) are the same, so we have: $$\frac{dx}{dt} = \alpha(p - x)^2,$$ with the same initial condition, \(x(0) = 0\). The logic for finding the limiting concentration is the same as in step 1: As the reaction progresses, the concentration of substances \(P\) and \(Q\) decreases, and so does the rate at which the reaction occurs (\(\alpha(p - x)^2\)). Therefore, for \(t \rightarrow \infty\), we have \(dx/dt = 0\). Solving for \(x\): $$0 = \alpha(p-x)^2$$ $$x_{lim} = p$$

Solve the initial value problem for p = q

Now we have the differential equation: $$\frac{dx}{dt} = \alpha(p - x)^2, \quad x(0) = 0$$ Rewriting the equation, we have: $$\frac{dx}{(p - x)^2} = \alpha dt$$ Now we integrate both sides: $$\int_{0}^{x}\frac{dy}{(p - y)^2} = \alpha \int_{0}^{t}ds$$ Integrating, we get: $$\left[-\frac{1}{p - y} \right]_0^x = \alpha t$$ Substituting the bounds of integration, we can find \(x(t)\): $$\frac{1}{p} - \frac{1}{p-x} = \alpha t$$ $$x(t) = p\left(1 - \frac{1}{1 + \alpha pt}\right)$$ Now we have the concentration of substance \(X\) at any time, \(x(t) = p\left(1 - \frac{1}{1 + \alpha pt}\right)\), for the case where substances \(P\) and \(Q\) are the same, and the limiting concentration as \(t \rightarrow \infty\) is \(p\).

Key concepts

Second Order Reaction

A second order reaction is a type of chemical reaction where the reaction rate is proportional to the product of the concentrations of two reactants. For example, in the reaction \(P + Q \rightarrow X\), the rate decreases as the concentrations of \(P\) and \(Q\) decrease. Second order reactions are common when two molecules must collide to form a product. This can be expressed with a rate equation of the form:

• \(\frac{dx}{dt} = \alpha(p-x)(q-x)\)

Here, \(\alpha\) is the rate constant, and \(p-x\) and \(q-x\) represent the concentrations of the reactants at time \(t\). The rate at which \(P\) and \(Q\) transform into \(X\) depends on their initial concentrations \(p\) and \(q\), as well as the amount already converted to \(X\), symbolized by \(x\). Understanding second order reactions is crucial for predicting product formation and determining reaction mechanisms.

Differential Equations in Chemistry

In chemistry, differential equations are often used to model the rates of reactions over time. They help us quantify how the concentration of reactants and products changes as a reaction progresses. The given differential equation for a second order reaction is:

• \(\frac{dx}{dt} = \alpha(p-x)(q-x)\)

This equation can be solved to understand how the concentration of product \(x(t)\) evolves. Solving it typically involves rearranging the equation and integrating both sides. For instance, you might rewrite it as:\(\frac{dx}{(p-x)(q-x)} = \alpha dt\)Then, separate the variables and integrate:\(\int \frac{1}{(p-y)(q-y)} dy = \alpha \int dt\)Such equations require techniques like partial fraction decomposition to integrate, as seen in the exercise. Differential equations allow chemists to create detailed models of reaction dynamics, offering deeper insights into reaction processes.

Rate of Reaction

The rate of a chemical reaction refers to how fast reactants are converted to products. It is often influenced by factors such as concentration, temperature, and catalysts. In the context of second order reactions, the rate is specifically dependent on the concentrations of two reactants. The reaction rate equation in this context is:

• \(\frac{dx}{dt} = \alpha(p-x)(q-x)\)

The rate typically decreases over time as the concentrations of \(P\) and \(Q\) decrease due to the formation of \(X\). Eventually, the rate becomes zero when the reaction reaches completion, meaning the maximum possible concentration of \(X\) has been achieved. Monitoring reaction rates is crucial for controlling industrial processes and enhancing the efficiency of chemical synthesis. By understanding how rates change over time, chemists can design experiments and processes that maximize yield and minimize unwanted byproducts.