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Cisplatin, [Pt(NH\_3)\_2Cl\_2], is used as an anti-cancer drug - AQA - A-Level Chemistry - Question 4 - 2020 - Paper 3

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Cisplatin, [Pt(NH\_3)\_2Cl\_2], is used as an anti-cancer drug. Cisplatin works by causing the death of rapidly dividing cells. Name the process that is prevented ... show full transcript

Worked Solution & Example Answer:Cisplatin, [Pt(NH\_3)\_2Cl\_2], is used as an anti-cancer drug - AQA - A-Level Chemistry - Question 4 - 2020 - Paper 3

Step 1

Name the process that is prevented by cisplatin during cell division.

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Answer

The process that is prevented by cisplatin during cell division is DNA replication.

Step 2

Give the equation for this reaction.

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Answer

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

Step 3

Complete Figure 1 to show how the platinum complex forms a cross-link between the guanine nucleotides.

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Answer

In Figure 1, draw a bond between the platinum atom in complex ion B and the nitrogen atoms of the adjacent guanine nucleotides. Indicate the cross-link by showing the two guanine nucleotides connected through the platinum.

Step 4

Explain how graphical methods can be used to process the measured results, to confirm that the reaction is first order.

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Answer

To confirm that the reaction is first-order, plot the concentration of cisplatin (y-axis) against time (x-axis). Calculate the gradients of the tangents to the curve at various points to determine the reaction rate. A straight line through the origin indicates a first-order reaction since the rate is directly proportional to the concentration.

Step 5

Complete Table 1.

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Answer

For T = 313 K, 1/T = 0.003194 and ln k = -14.9. For T = 318 K, 1/T = 0.003141 and ln k = -13.6.

Step 6

Calculate the activation energy, E_a, in kJ mol^{-1}.

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Answer

Using the Arrhenius equation, the gradient of the line plotted on the graph of ln k against 1/T can be used to calculate the activation energy. Given that the gradient = -E_a/R, we substitute the gradient (calculated from the graph) into the equation to find ( E_a = -\text{gradient} \times R ). For a gradient of -13.125, ( E_a = -(-13.125) \times 8.31 \approx 109.09 \text{ kJ mol}^{-1} ).

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