Chemical reaction kinetics

7 In a certain chemical process a substance is being formed, and \(t\) minutes after the start of the process there are \(m\) grams of the substance present. In the process the rate of increase of \(m\) is proportional to \(( 50 - m ) ^ { 2 }\). When \(t = 0 , m = 0\) and \(\frac { \mathrm { d } m } { \mathrm {~d} t } = 5\).

1. Show that \(m\) satisfies the differential equation $$\frac { \mathrm { d } m } { \mathrm {~d} t } = 0.002 ( 50 - m ) ^ { 2 }$$
2. Solve the differential equation, and show that the solution can be expressed in the form $$m = 50 - \frac { 500 } { t + 10 }$$
3. Calculate the mass of the substance when \(t = 10\), and find the time taken for the mass to increase from 0 to 45 grams.
4. State what happens to the mass of the substance as \(t\) becomes very large.

5 In a certain chemical process a substance \(A\) reacts with another substance \(B\). The masses in grams of \(A\) and \(B\) present at time \(t\) seconds after the start of the process are \(x\) and \(y\) respectively. It is given that \(\frac { \mathrm { d } y } { \mathrm {~d} t } = - 0.6 x y\) and \(x = 5 \mathrm { e } ^ { - 3 t }\). When \(t = 0 , y = 70\).

1. Form a differential equation in \(y\) and \(t\). Solve this differential equation and obtain an expression for \(y\) in terms of \(t\).
2. The percentage of the initial mass of \(B\) remaining at time \(t\) is denoted by \(p\). Find the exact value approached by \(p\) as \(t\) becomes large.

2 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?

1 A drug is administered by an intravenous drip. The concentration, \(x\), of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after \(t\) hours is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right)$$ where \(0 \leqslant x < 1\), and \(k\) is a positive constant. Initially, \(x = 0\).

1. Express \(\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }\) in partial fractions.
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2. Hence solve the differential equation to show that \(\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }\).
3. After 1 hour the drug concentration reaches \(75 \%\) of its maximum value and so \(x = 0.75\). Find the value of \(k\), and the time taken for the drug concentration to reach \(90 \%\) of its maximum value.
4. Rearrange the equation in part (ii) to show that \(x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }\). Verify that in the long term the drug concentration approaches its maximum value.