This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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--- QUESTION 4 NOV 2014 4.1 The boiling point is the temperature at which the vapor pressure of a liquid equals the external atmospheric pressure, causing the liquid to change into a gas. Temperature at which vapor pressure equals external pressure 4.2 Fully explain the difference in boiling points between compound P and: 4.2.1 Compound Q (NH_3) Step 1: Identify the intermolecular forces. Compound P (CH_4) is a non-polar molecule and only experiences weak London dispersion forces. Compound Q (NH_3) is a polar molecule and can form strong hydrogen bonds, in addition to dipole-dipole forces and London dispersion forces. Step 2: Compare the strength of intermolecular forces. Hydrogen bonds are significantly stronger intermolecular forces than London dispersion forces. Step 3: Relate intermolecular forces to boiling point. More energy is required to overcome the strong hydrogen bonds in NH_3 compared to the weak London dispersion forces in CH_4. Therefore, NH_3 has a higher boiling point (-33 °C) than CH_4 (-164 °C). 4.2.2 Compound S (SiH_4) Step 1: Identify the intermolecular forces. Both compound P (CH_4) and compound S (SiH_4) are non-polar molecules with tetrahedral shapes, meaning they only experience London dispersion forces. Step 2: Compare molecular masses. The molecular mass of CH_4 is approximately 16 g/mol, while the molecular mass of SiH_4 is approximately 32 g/mol. SiH_4 has a higher molecular mass and more electrons. Step 3: Relate molecular mass to the strength of London dispersion forces. As the molecular mass and number of electrons increase, the electron cloud becomes more polarizable, leading to stronger London dispersion forces between molecules. Step 4: Relate intermolecular forces to boiling point. More energy is required to overcome the stronger London dispersion forces in SiH_4 compared to CH_4. Therefore, SiH_4 has a higher boiling point (-112 °C) than CH_4 (-164 °C). 4.3 Explain why the boiling points of compounds Q (NH_3) and R (H_2O) differ by referring to ELECTRONEGATIVITY and DEGREE OF POLARITY: Step 1: Compare boiling points and identify intermolecular forces. Compound R (H_2O) has a boiling point of 100 °C, while compound Q (NH_3) has a boiling point of -33 °C. Both molecules are polar and can form hydrogen bonds. Step 2: Compare electronegativity and bond polarity. Oxygen is more electronegative than nitrogen. This makes the O-H bonds in H_2O more polar than the N-H bonds in NH_3. Step 3: Compare the extent and strength of hydrogen bonding. Due to the greater electronegativity of oxygen and the presence of two lone pairs, H_2O can form more and stronger hydrogen bonds per molecule (up to four) compared to NH_3 (which typically forms fewer and weaker hydrogen bonds). Step 4: Conclude the effect on boiling point. The stronger and more extensive network of hydrogen bonds in H_2O requires significantly more energy to overcome during boiling, resulting in a much higher boiling point than NH_3. 4.4 Write down the letter from the table that represents the following: 4.4.1 ONE polar compound: Q (or R) 4.4.2 ONE non-polar compound: P (or S) --- QUESTION 3 NOV 2015 3.1 The boiling point is the temperature at which the vapor pressure of a liquid equals the external atmospheric pressure, causing the liquid to change into a gas. Temperature at which vapor pressure equals external pressure 3.2 Calculate the molecular mass of compound D (C_9H_20): Step 1: Write down the formula for molecular mass. Molecular mass = (number of C atoms × atomic mass of C) + (number of H atoms × atomic mass of H) Step 2: Substitute atomic masses. Atomic mass of C ≈ 12.01 g/mol Atomic mass of H ≈ 1.01 g/mol Molecular mass of C_9H_20 = (9 × 12.01) + (20 × 1.01) Step 3: Calculate the molecular mass. Molecular mass = 108.09 + 20.20 = 128.29 g/mol 128.29 g/mol 3.3 In what phase is compound B at 25 °C? Compound B (C_2H_6) has a boiling point of -89 °C. Since 25 °C is above its boiling point, compound B will be in the gaseous phase. Gas 3.4 Name the type of intermolecular force present in compound A (CH_4): Compound A (CH_4) is a non-polar molecule. London dispersion forces (or Van der Waals forces) 3.5 Explain why the boiling point increases from compound A to E: Step 1: Identify the trend in molecular structure. Compounds A to E are alkanes with increasing chain length and molecular mass (CH_4, C_2H_6, C_7H_16, C_9H_20, C_20H_42). Step 2: Relate molecular mass to intermolecular forces. As the molecular mass increases, the number of electrons in the molecule increases. This leads to a greater polarizability of the electron cloud, resulting in stronger London dispersion forces between molecules. Step 3: Conclude the effect on boiling point. More energy is required to overcome these stronger London dispersion forces, causing the boiling point to increase from compound A to E. 3.6 How does the vapour pressure of compound B compare to the vapour pressure of compound C? Write down only HIGHER THAN, LOWER THAN or EQUAL TO. Compound B has a boiling point of -89 °C. Compound C has a boiling point of 36 °C. A lower boiling point indicates weaker intermolecular forces, which means molecules can escape into the gas phase more easily, leading to a higher vapor pressure. Therefore, compound B will have a higher vapor pressure than compound C. HIGHER THAN Drop the next question.