在化学反应中，化合物 X 和 Y 反应生成一种产物。设 $x$ 和 $y$ 分别表示在实验开始后 $t$ 分钟时，$X$ 和 $Y$ 的浓度，单位为摩尔每千升（mol/kL）。 $X$ 和 $Y$ 的初始浓度分别由 $x_0$ 和 $y_0$ mol/kL 给出。

1. Separate the variables $x$ and $t$, and set up the integrals.

$\frac{dx}{dt} = -ax$
$\int \frac{1}{x} \, dx = \int -a \, dt$

2. Integrate both sides to obtain the general solution.

$\ln x = -at + C$
(Since $x$ represents concentration, $x > 0$ and the absolute value can be omitted.)

3. Apply the initial condition to find the constant of integration.

Given that the initial concentration of $X$ is $x_0$, when $t = 0$, $x = x_0$.
$\ln x_0 = -a(0) + C$
$C = \ln x_0$

4. Rearrange the equation to express $x$ in terms of $t$, $x_0$ and $a$.

$\ln x = -at + \ln x_0$
$\ln x - \ln x_0 = -at$
$\ln \left(\frac{x}{x_0}\right) = -at$
$\frac{x}{x_0} = e^{-at}$
$x = x_0 e^{-at}$

1. Set up the equation using the given definition of half-life and the solution from part (a)(i).

From part (a)(i), the solution to the differential equation is $x = x_0 e^{-at}$.

Given that $t_{0.5}$ is the time taken for $x$ to reach half its initial value ($x_0$), substitute $x = \frac{x_0}{2}$ and $t = t_{0.5}$ into the equation:
$\frac{x_0}{2} = x_0 e^{-a t_{0.5}}$

2. Simplify and solve the resulting exponential equation for $t_{0.5}$.

Divide both sides by $x_0$:
$\frac{1}{2} = e^{-a t_{0.5}}$

Take the natural logarithm on both sides:
$\ln\left(\frac{1}{2}\right) = -a t_{0.5}$
$-\ln 2 = -a t_{0.5}$

$t_{0.5} = \frac{\ln 2}{a} \text{ (shown)}$

1. Express $y$ in terms of $x$ using the initial conditions.

Given the equation $y_0 - y = 2(x_0 - x)$, and the initial concentrations $x_0 = 1$ and $y_0 = 4$:
$4 - y = 2(1 - x)$
$y = 4 - 2(1 - x)$
$y = 2 + 2x = 2(x + 1)$

2. Formulate the differential equation and substitute $y$ to obtain the final equation.

The rate of the reaction (the rate of consumption of $X$) is proportional to $xy^2$. Since the concentration of $X$ is decreasing, its rate of change is negative:
$\frac{dx}{dt} = -k x y^2 \quad \text{for some constant } k > 0$
Substituting $y = 2(x+1)$ into the equation:
$\frac{dx}{dt} = -k x [2(x+1)]^2$
$\frac{dx}{dt} = -4k x (x+1)^2$
Letting $b = 4k$, which is a positive constant since $k > 0$, we have:
$\frac{dx}{dt} = -bx(x+1)^2 \quad \text{(Shown)}$

1. Separate variables and perform partial fraction decomposition on the integrand.

$\frac{\mathrm{d}x}{\mathrm{d}t} = -bx(x+1)^2$
$\int \frac{1}{x(x+1)^2} \,\mathrm{d}x = \int -b \,\mathrm{d}t$
Let $\frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$.
$1 = A(x+1)^2 + Bx(x+1) + Cx$
Substituting $x = 0$ gives $A = 1$.
Substituting $x = -1$ gives $C = -1$.
Comparing coefficients of $x^2$: $A + B = 0 \implies B = -1$.
Thus, $\frac{1}{x(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}$.

2. Integrate both sides to obtain the general solution.

$\int \left( \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) \mathrm{d}x = \int -b \,\mathrm{d}t$
$\ln|x| - \ln|x+1| + \frac{1}{x+1} = -bt + C$
Since concentration $x > 0$, this simplifies to:
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = -bt + C$

3. Use the initial condition $x = 1$ at $t = 0$ to find the constant of integration $C$.

From the problem context, the initial concentration of $X$ is $1 \text{ mol/kL}$, so $x = 1$ when $t = 0$.
Substituting $t = 0$ and $x = 1$:
$\ln\left(\frac{1}{2}\right) + \frac{1}{2} = -b(0) + C$
$C = \frac{1}{2} - \ln 2$
So the equation becomes:
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = -bt + \frac{1}{2} - \ln 2$

4. Use the condition $x = 0.5$ at $t = 1$ to determine the constant $b$.

Given $x = 0.5$ when $t = 1$:
$\ln\left(\frac{0.5}{1.5}\right) + \frac{1}{1.5} = -b(1) + \frac{1}{2} - \ln 2$
$\ln\left(\frac{1}{3}\right) + \frac{2}{3} = -b + \frac{1}{2} - \ln 2$
$b = \frac{1}{2} - \ln 2 - \ln\left(\frac{1}{3}\right) - \frac{2}{3}$
$b = \ln 3 - \ln 2 - \frac{1}{6} = \ln\left(\frac{3}{2}\right) - \frac{1}{6}$

5. Substitute $t = 2$ and solve the resulting non-linear equation for $x$ using a graphing calculator.

Substitute $t = 2$ and the value of $b$:
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = -2\left(\ln\left(\frac{3}{2}\right) - \frac{1}{6}\right) + \frac{1}{2} - \ln 2$
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = -2\ln\left(\frac{3}{2}\right) + \frac{1}{3} + \frac{1}{2} - \ln 2$
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = \frac{5}{6} - \ln\left(\frac{9}{4} \times 2\right) = \frac{5}{6} - \ln\left(\frac{9}{2}\right)$
$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = \frac{5}{6} + \ln\left(\frac{2}{9}\right) \approx -0.67074$
Using a graphing calculator to solve for $x > 0$:
$x \approx 0.31383 \dots$
$x = 0.314 \text{ (to 3 s.f.)}$