2025 ASRJC P1 Q11

2025 ASRJC P1 Q11

13 marks

In a chemical reaction, compounds \(X\) and \(Y\) react to form a product. Let \(x\) and \(y\) denote the concentrations, in mol per kilolitres (mol/kL), of \(X\) and \(Y\) respectively, at time \(t\) minutes after the start of the experiment.

The initial concentrations of \(X\) and \(Y\) are given by \(x_0\) and \(y_0\) mol/kL.

  1. In a particular experiment where \(Y\) is present in excess, the reaction rate can be modeled as a pseudo-first-order reaction, leading to the differential equation \(\frac{\mathrm{d}x}{\mathrm{d}t} = -ax,\) where \(a\) is a positive constant.

    1. Solve this differential equation, expressing \(x\) in terms of \(t\), \(x_0\) and \(a\).[3]
    2. The half-life of the reaction, denoted by \(t_{0.5}\), is defined as the time taken for the concentration of \(X\) to decrease to half its initial value. Show that \(t_{0.5} = \frac{\ln 2}{a}\).[2]
  2. In another experiment conducted by a chemist, the rate of the reaction is directly proportional to the product of the concentration of \(X\) and the square of the concentration of \(Y\). It is also known that at any instance during the reaction, every \(1\) mol of \(X\) reacts with every \(2\) mol of \(Y\), giving the equation \(y_0 - y = 2(x_0 - x)\).

The initial concentrations of \(X\) and \(Y\) are \(1\) mol/kL and \(4\) mol/kL respectively.

    1. Show that \(\frac{\mathrm{d}x}{\mathrm{d}t} = -bx(x + 1)^2,\) where \(b\) is a positive constant.

      [2]
    2. It is given that the concentration of \(X\) is \(0.5\) mol/kL at the instance \(1\) min after the start of the experiment. Find the concentration of \(X\) at the instance \(2\) min after the start of the experiment, giving your answer to \(3\) significant figures.[6]

Solution:

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Answer:(a)(i) \(x=x_0e^{-at}\). (ii) \(t_{0.5}=\frac{\ln2}{a}\). (b)(i) \(\frac{\mathrm dx}{\mathrm dt}=-bx(x+1)^2\), \(b=4k>0\). (ii) \(0.314\) mol kL\(^{-1}\).

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