Naphthalene, C₁₀H₈, is an aromatic organic compound that is a solid at room temperature and pressure. The enthalpy change of fusion, ∆Hfusion, of naphthalene is the enthalpy change that occurs when naphthalene melts. It cannot be directly measured. C₁₀H₈(s) ⇌ C₁₀H₈(l) The melting point of naphthalene depends on its purity. When other compounds dissolve in liquid naphthalene its melting point is lowered. Diphenylamine, (C₆H₅)₂NH, dissolves in liquid naphthalene. By varying the amount of diphenylamine and measuring the melting points of the different mixtures, the enthalpy change of fusion of naphthalene can be calculated. (a) The equation that links the melting point of a naphthalene-diphenylamine mixture with its enthalpy change of fusion, ∆Hfusion, is shown. log Y = A - ∆Hfusion / (2.30 x RTm) Y = mole fraction of naphthalene R = molar gas constant, 8.31 J K⁻¹ mol⁻¹ Tₘ = melting point of naphthalene in K A is a constant

1. (a)(i)A series of experiments is performed using the same amount, 0.100 mol, of naphthalene each time. Calculate the mass of naphthalene, C₁₀H₈, that should be used for each of these experiments. [Aᵣ: C, 12.0; H, 1.0][1]
2. (a)(ii)The melting point and freezing point of a substance are the same. The melting point, Tₘ, of a substance can be found by recording the temperature at which the substance freezes, measured when crystals first start to appear on cooling. The results of a series of experiments using 0.100 mol of naphthalene and different masses of diphenylamine are shown. Process the results to complete the table. Record all your data to three significant figures. The mole fraction of naphthalene, Y, is calculated as shown. Y = nN / (nN + nD) nN = amount in moles of naphthalene, C₁₀H₈, = 0.100 nD = amount in moles of diphenylamine, (C₆H₅)₂NH amount of (C₆H₅)₂NH, nD /mol | mole fraction of C₁₀H₈, Y | temperature at which crystals appear, Tₘ /K | 1/Tₘ /10⁻³K⁻¹ | log Y ---|---|---|---|--- 0.00 | 1.00 | 353 | 2.83 | 0.00 0.00888 | | 349 | | 0.0178 | | 345 | | 0.0266 | | 341 | | 0.0355 | | 338 | | 0.0444 | | 334 | | 0.0533 | | 331 | | 0.0621 | | 329 | | 0.0769 | | 325 | |[3]
3. (b)Plot a graph on the grid to show the relationship between 1/Tₘ and log Y. Use a cross (x) to plot each data point. Draw the straight line of best fit. [Figure showing graph grid][2]
4. (c)(i)Use the graph to determine the gradient of the line of best fit. State the co-ordinates of both points you used in your calculation.[2]
5. (c)(ii)Use your answer to (c)(i) to determine the value of the enthalpy change of fusion of naphthalene, ∆Hfusion, in kJ mol⁻¹.[1]
6. (d)(i)Do you consider the results obtained to be reliable? Explain your answer.[1]
7. (d)(ii)Different literature values for the enthalpy change of fusion of naphthalene suggest that 10.00 g of naphthalene require between 1.45 kJ and 1.47 kJ to melt. Use this information to calculate the range of ∆Hfusion values of naphthalene, C₁₀H₈, given in literature. Use your values to comment on the accuracy of the experimental procedure. [Aᵣ: C, 12.0; H, 1.0] If you were not able to calculate ∆Hfusion in (c)(ii), you may use 18.4 kJ mol⁻¹, but this may not be the correct answer.[2]
8. (e)The enthalpy change calculated in this reaction is actually ∆H₁, shown in the Hess' cycle. [Figure showing Hess' cycle: C₁₀H₈(s) --∆Hfusion--> C₁₀H₈(l) --∆Hmixing--> C₁₀H₈(l) mixed with diphenylamine. There's also an arrow from C₁₀H₈(s) to C₁₀H₈(l) mixed with diphenylamine labelled ∆H₁.] It is assumed that the enthalpy change when C₁₀H₈(l) and diphenylamine are mixed, ∆Hmixing, is zero, and therefore ∆H₁ = ∆Hfusion. State how the value of ∆H₁ compares to the value of ∆Hfusion if the mixing of naphthalene and diphenylamine is endothermic. Explain your answer.[2]
9. (f)(i)A student incorrectly uses a value for the Mᵣ of diphenylamine that is too low. This produces incorrect values for the mole fraction, Y. Y = nN / (nN + nD) nN = amount in moles of naphthalene nD = amount in moles of diphenylamine Predict the effect this will have on the calculated values of Y. Explain your answer.[1]
10. (f)(ii)The student uses the incorrectly calculated value of Y from (f)(i) in the determination of ∆Hfusion log Y = A - ∆Hfusion / (2.30 x RTm) Predict how the student's calculated value of ∆Hfusion is different from the actual value. Explain your answer.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.