Consider the following heating curve for Substance X (with a molar mass of 45g/m

Question

Consider the following heating curve for Substance X (with a molar mass of 45g/mol). A 10.0 g sample of Substance X is heated at a constant rate of 4.56 kJ/min and the temperature measured every minute. The results of the experiment is shown below 6. 100 "C I1 a. Identify the melting point and the boiling point of substance X7 b. Why does the temperature at portion 2 in the graph remain constant? c. Assuming there is no heat loss and all the added heat is absorbed by the substance estimate the enthalpy of vaporization of substance X in kJ/mor? Using the data from the graph, deduce the relative strength of intermolecular forces in substance X compared to water. Justify your answer d.

Explanation / Answer

Part a

Step 2 is a melting (phase change) process where the temperature remains constant.

Melting point Tm = 20 °C

Step 4 is a boiling (phase change) process where the temperature remains constant.

Boiling point Tb = 120 °C

Part b

Step 1 shows that the solid is heat from 0 to 20 °C

Step 2 shows the phase change occurs at 20 ° C (constant temperature) by absorbing the latent heat of fusion. This latent heat is used to break the bonds in solid phase and convert it into liquid phase. Kinetic energy of molecules remains same so the temperature.

Part c

Moles of X = mass/molecular weight

= 10g / 45g/mol

= 0.222 mol

Heated at a constant rate = 4.56 kJ/min

Time taken to vaporization = 20 - 6.5 = 13.5 min

Enthalpy of vaporization = (4.56 kJ/min x 13.5 min) / 0.222 mol

= 277.297 kJ/mol

Part d

The strength of intermolecular forces depends on the latent heat of fusion and latent heat of vaporization of the substance.

If heat of fusion and latent heat of vaporization of X required to break the bonds are higher than that of in water. The relative strength of intermolecular forces will be higher.

Heated at a constant rate = 4.56 kJ/min

Time taken to melting = 4 - 2 = 2 min

Heat of fusion = (4.56 kJ/min x 2 min) / 0.222 mol

= 41.08 kJ/mol

For water heat of fusion = 6.02 kJ/mol

Heat of fusion of X (41.08) > heat of fusion of water(6.02)

Heat of vaporization of X(277.297) > heat of vaporization of water (40.7)

Relative strength of intermolecular forces in substance X is much higher than that in water.

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