| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Differential equations |
| Type | Chemical reaction kinetics |
| Difficulty | Standard +0.3 This is a standard separable differential equations question requiring partial fractions, integration, and manipulation of logarithms—all routine C4 techniques. Part (a) follows a predictable method, part (b) applies initial conditions straightforwardly, and part (c) requires simple limit evaluation. While multi-step, it demands no novel insight and is slightly easier than the average A-level question due to its formulaic nature. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations |
In a chemical reaction two substances combine to form a third substance. At time $t$, $t \geq 0$, the concentration of this third substance is $x$ and the reaction is modelled by the differential equation
$$\frac{dx}{dt} = k(1 - 2x)(1 - 4x), \text{ where } k \text{ is a positive constant.}$$
\begin{enumerate}[label=(\alph*)]
\item Solve this differential equation and hence show that
$$\ln\left|\frac{1 - 2x}{1 - 4x}\right| = 2kt + c, \text{ where } c \text{ is an arbitrary constant.}$$ [7]
\item Given that $x = 0$ when $t = 0$, find an expression for $x$ in terms of $k$ and $t$. [4]
\item Find the limiting value of the concentration $x$ as $t$ becomes very large. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [13]}}