Chemical reaction kinetics

4. During a chemical reaction, a compound is being made from two other substances. At time \(t\) hours after the start of the reaction, \(x \mathrm {~g}\) of the compound has been produced. Assuming that \(x = 0\) initially, and that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$

1. show that it takes approximately 7 minutes to produce 2 g of the compound.
2. Explain why it is not possible to produce 3 g of the compound.
   4. continued

In a certain chemical process a substance \(A\) reacts with and reduces a substance \(B\). The masses of \(A\) and \(B\) at time \(t\) after the start of the process are \(x\) and \(y\) respectively. It is given that \(\frac{dy}{dt} = -0.2xy\) and \(x = \frac{10}{(1 + t)^2}\). At the beginning of the process \(y = 100\).

1. Form a differential equation in \(y\) and \(t\), and solve this differential equation. [6]
2. Find the exact value approached by the mass of \(B\) as \(t\) becomes large. State what happens to the mass of \(A\) as \(t\) becomes large. [2]

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 $$\frac{dx}{dt} = k(1 - 2x)(1 - 4x), \text{ where } k \text{ is a positive constant.}$$

1. 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]
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\). [4]
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. [2]

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 $$\frac{dx}{dt} = k(1 - 2x)(1 - 4x), \text{ where } k \text{ is a positive constant.}$$

1. 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]
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\). [4]
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. [2]

In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = \frac{M}{t(1+t^2)}.$$

1. Find \(\int \frac{t}{1+t^2} dt\). [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1+t^2)} = \frac{A}{t} + \frac{Bt+C}{1+t^2}.$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac{Kt}{\sqrt{1+t^2}},$$ where \(K\) is a constant. [6]
4. When \(t = 1\), \(M = 25\). Calculate \(K\). What is the mass of the chemical in the long term? [4]

In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = -\frac{M}{2(1 + \frac{t}{2})}$$

1. Find \(\int \frac{1}{1 + \frac{t}{2}} dt\) [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1 + \frac{t}{2})} = \frac{A}{t} + \frac{Bt + C}{1 + \frac{t}{2}}$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M \sim \frac{K}{.1 + \frac{t}{2}}$$ where \(K\) is a constant. [6]
4. When \(t = 1\), \(M = 25\). Calculate \(K\) What is the mass of the chemical in the long term? [4]