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

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Cisplatin, [Pt(NH₃)₂Cl₂], is used as an anti-cancer drug. Cisplatin works by causing the death of rapidly dividing cells. Name the process that is prevented by cis... show full transcript

Worked Solution & Example Answer:Cisplatin, [Pt(NH₃)₂Cl₂], 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(NH₃)₂Cl₂] + H₂O → [Pt(NH₃)₂(H₂O)Cl]⁺ + Cl⁻

Step 3

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

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Answer

In Figure 1, draw a line connecting the platinum atom to the nitrogen atoms (N) of the two guanine nucleotides to illustrate the cross-link. Label the connection as a coordination bond.

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, one should plot the concentration of cisplatin against time on the y-axis. The slope of the tangent lines at various points provides the rate at which concentration changes. If the plot yields a straight line when plotting ln[concentration] vs. time, it indicates the reaction is first order.

Step 5

Complete Table 1.

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Answer

The values for 1/T and ln k are filled in based on the given data. The rate constant, k, is calculated using the specified temperatures, resulting in ln k values that correspond to the respective temperatures.

Step 6

Calculate the activation energy, Eₐ, in kJ mol⁻¹.

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Answer

To calculate the activation energy, Eₐ, use the Arrhenius equation. First, determine the gradient of the ln k vs. 1/T graph, which equals -Eₐ/R. The gradient is found to be approximately -13.125. Using the gas constant R = 8.31 J K⁻¹ mol⁻¹, the activation energy is calculated as follows:

Ea=extgradientimesR=(13.125)imes8.31Eₐ = - ext{gradient} imes R = -(-13.125) imes 8.31

Calculating this gives:

Ea=109.09extkJmol1Eₐ = 109.09 ext{ kJ mol}^{-1}

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