Cisplatin, [Pt(NH3)2Cl2], is used as an anti-cancer drug - AQA - A-Level Chemistry - Question 4 - 2020 - Paper 3

The equation for the reaction when cisplatin reacts with water to form a complex ion B can be represented as:

$[Pt(NH_3)_2Cl_2] + H_2O \rightarrow [Pt(NH_3)_2(H_2O)]^{2+} + Cl^-$

In Figure 1, indicate the cross-link formation by drawing a bond (such as a dotted line) connecting the nitrogen atoms of the two guanine nucleotides with a platinum atom at the center.

To confirm that the reaction is first order, we can plot the concentration of cisplatin (y-axis) against time (x-axis) to calculate the rate at different intervals. Tangents to the curve can be plotted, and if the rates are directly proportional to the concentrations of cisplatin, this would indicate a first-order reaction. Additionally, a plot of ln(concentration) against time should yield a straight line if the reaction is first order.

The missing values for the rate constant in Table 1 can be filled based on the corresponding temperature values. For example:

• For 293 K, the rate constant is 1.97 x 10^-6 s^-1, which results in ln k = -17.7.
• Continue filling in the values accordingly.

To calculate the activation energy, you can use the slope of the ln k versus 1/T graph. From the graph (Figure 2), the gradient (m) is -Ea/R. Thus, rearranging gives Ea = -gradient × R.

Using the gas constant R = 8.31 J K–1 mol–1 and converting the gradient from the plot:

• If gradient = -13.125, then:

$Ea = -(-13.125) \times 8.31 \approx 109.09 \text{ kJ mol}^{-1}$

Therefore, the activation energy Ea is approximately 109 kJ mol–1.