Question
The element sulfur has many industrial uses.
(a) (i) Determine the standard enthalpy of reaction (\(ΔH^{\theta}_r\)), in kJ \(mol^{-1}\), for the oxidation of \(SO_2\) to \(SO_3\).
▶️Answer/Explanation
Ans:
The correct equation for the standard enthalpy of reaction (\(ΔH^\theta_r\)) is:
\[ ΔH^\theta_r = ΔH^\circ_f(SO_3) – ΔH^\circ_f(SO_2) – ΔH^\circ_f(O_2) \]
Now, substituting the given values:
\[ ΔH^\theta_r = (-395.8 \, \text{kJ/mol}) – (-296.8 \, \text{kJ/mol}) – 0 \]
\[ ΔH^\theta_r = -395.8 \, \text{kJ/mol} +296.8 \, \text{kJ/mol} \]
\[ ΔH^\theta_r = -99 \, \text{kJ/mol} \]
Therefore, the correct standard enthalpy of reaction (\(ΔH^\theta_r\)) for the oxidation of \(SO_2\) to \(SO_3\) is \(-99 \, \text{kJ/mol}\).
Question
(ii) Formulate equations showing how \(SO_2\) and \(SO_3\) lead to acid deposition.
\(SO_2:\) …………………………………..
\(SO_3:\) …………………………………..
▶️Answer/Explanation
Ans:
The formation of acid deposition from \(SO_2\) and \(SO_3\) involves their reactions with atmospheric water to produce sulfuric acid (\(H_2SO_4\)), which contributes to acid rain. Here are the equations representing these processes:
1. \(SO_2\) leading to acid deposition:
\[ SO_2(g) + \frac{1}{2} O_2(g) + H_2O(l) \rightarrow H_2SO_4(aq) \]
\[ \text{Sulfur dioxide} + \text{Oxygen} + \text{Water} \rightarrow \text{Sulfuric acid} \]
2. \(SO_3\) leading to acid deposition:
\[ SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq) \]
\[ \text{Sulfur trioxide} + \text{Water} \rightarrow \text{Sulfuric acid} \]
In these reactions, sulfur dioxide (\(SO_2\)) or sulfur trioxide (\(SO_3\)) reacts with water (\(H_2O\)) in the atmosphere to form sulfuric acid (\(H_2SO_4\)), a strong acid. The resulting sulfuric acid can then contribute to acid deposition when it falls to the Earth’s surface as acid rain, snow, fog, or dry particles.
Question
(iii) Explain the polarity of the S—O bond. Use section 8 of the data booklet.
▶️Answer/Explanation
Ans:
The polarity of a chemical bond is determined by the difference in electronegativity between the atoms involved in the bond. Electronegativity is a measure of an atom’s ability to attract electrons towards itself in a chemical bond. The greater the difference in electronegativity between two atoms, the more polar the bond between them.
In the case of the sulfur-oxygen (\(S-O\)) bond in sulfur dioxide (\(SO_2\)) and sulfur trioxide (\(SO_3\)), sulfur is less electronegative than oxygen. Oxygen is one of the most electronegative elements, while sulfur is less electronegative. As a result, oxygen attracts the shared electrons in the \(S-O\) bond more strongly towards itself, leading to a partial negative charge (\(δ^-\)) on the oxygen atom and a partial positive charge (\(δ^+\)) on the sulfur atom.
This unequal sharing of electrons creates a dipole moment in the \(S-O\) bond, with the oxygen end being slightly negative and the sulfur end being slightly positive. Therefore, the \(S-O\) bond is polar due to the difference in electronegativity between sulfur and oxygen.
It’s worth noting that the polarity of the \(S-O\) bond contributes to the overall polarity of molecules containing \(SO_2\) and \(SO_3\), such as sulfuric acid (\(H_2SO_4\)), which is a polar molecule due to the presence of multiple polar \(S-O\) bonds.
Question
(b) The combustion of 0.1 moles of sulfur (S) was demonstrated in a school laboratory using the following apparatus in a fume cupboard.
(i) Calculate the enthalpy of combustion of sulfur, \(ΔH_c\), in kJ \(mol^{-1}\) from this data.
Use sections 1 and 2 of the data booklet.
▶️Answer/Explanation
Ans:
To calculate the enthalpy of combustion of sulfur (\(ΔH_c\)), we can use the formula:
\[ q = m \cdot c \cdot ΔT \]
where:
\( q \) is the heat absorbed or released (in joules),
\( m \) is the mass of the substance (in grams),
\( c \) is the specific heat capacity of water (4.18 J/g°C),
\( ΔT \) is the change in temperature (in °C).
First, let’s calculate the heat absorbed by the water:
\[ q = m \cdot c \cdot ΔT \]
Given:
Mass of water (\( m \)) = 50.00 g,
Specific heat capacity of water (\( c \)) = 4.18 J/g°C,
Change in temperature (\( ΔT \)) = \( 35.0°C – 20.0°C = 15.0°C \).
Substitute the given values into the equation:
\[ q = 50.00 \, \text{g} \times 4.18 \, \text{J/g°C} \times 15.0 \, \text{°C} \]
\[ q = 3135 \, \text{J} \]
Next, let’s convert the heat absorbed by the water from joules to kilojoules:
\[ q = 3135 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} \]
\[ q = 3.135 \, \text{kJ} \]
Now, let’s calculate the moles of sulfur (\( n \)) that were combusted:
Given:
Moles of sulfur (\( n \)) = 0.1 moles.
The enthalpy change for the combustion of sulfur (\( ΔH_c \)) can be calculated using the equation:
\[ ΔH_c = \frac{q}{n} \]
Substitute the values:
\[ ΔH_c = \frac{3.135 \, \text{kJ}}{0.1 \, \text{mol}} \]
\[ ΔH_c = 31.35 \, \text{kJ/mol} \]
Therefore, the enthalpy of combustion of sulfur (\( ΔH_c \)) is \( 31.35 \, \text{kJ/mol} \).
Question
(ii) Suggest the major source of systematic error in this experiment and an improvement to reduce this error.
Source of systematic error: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Improvement: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
▶️Answer/Explanation
Ans:
Source of systematic error: One major source of systematic error in this experiment could be heat loss to the surroundings during the combustion process. Since the experiment is conducted in a fume cupboard rather than in a closed system, heat from the combustion reaction may escape to the surroundings, leading to an underestimation of the heat absorbed by the water.
Improvement: To reduce this error, an improvement would be to conduct the experiment in a closed system where heat loss to the surroundings is minimized. For example, the apparatus could be modified to include a well-insulated container surrounding the reaction vessel and water to minimize heat loss. Additionally, using a calorimeter with better insulation and ensuring a tight seal around the reaction vessel can help minimize heat loss and improve the accuracy of the experiment.
Question
▶️Answer/Explanation
Ans:
To calculate the percentage uncertainty in the temperature change, we need to consider the absolute uncertainty in the temperature change and the measured value of the temperature change.
Given:
Initial temperature of water (\(T_{\text{initial}}\)) = \(20.0^\circ \text{C}\),
Final temperature of water (\(T_{\text{final}}\)) = \(35.0^\circ \text{C}\),
Absolute uncertainty in temperature measurement = \(0.5^\circ \text{C}\).
The temperature change (\(ΔT\)) is calculated as:
\[ ΔT = T_{\text{final}} – T_{\text{initial}} = 35.0^\circ \text{C} – 20.0^\circ \text{C} = 15.0^\circ \text{C} \]
The absolute uncertainty in the temperature change (\(ΔT\)) is the sum of the absolute uncertainties in the initial and final temperatures:
\[ \text{Absolute uncertainty in } ΔT = \text{Absolute uncertainty in } T_{\text{initial}} + \text{Absolute uncertainty in } T_{\text{final}} = 0.5^\circ \text{C} + 0.5^\circ \text{C} = 1.0^\circ \text{C} \]
\[ \text{Percentage uncertainty in } ΔT = \frac{\text{Absolute uncertainty in } ΔT}{|ΔT|} \times 100\% \]
\[ \text{Percentage uncertainty in } ΔT = \frac{1.0^\circ \text{C}}{15.0^\circ \text{C}} \times 100\% \]
\[ \text{Percentage uncertainty in } ΔT \approx 6.7\% \]
Therefore, the percentage uncertainty in the temperature change is approximately \(6.7\%\) to two significant figures.
Question
(iv) Suggest one way of reducing the percentage uncertainty in this experiment.
▶️Answer/Explanation
Ans:
One way to reduce the percentage uncertainty in this experiment is to use a more precise temperature measuring device. For example, instead of using a thermometer with a scale in increments of 1°C, a digital thermometer with a higher precision scale (e.g., 0.1°C or 0.01°C) could be used. This would allow for more accurate measurement of the initial and final temperatures of the water, reducing the absolute uncertainty in the temperature change and thus reducing the percentage uncertainty in the experiment.
Question
(v) Calculate the overall percentage error of this experiment. Use part (b)(i) and section 13 of the data booklet. (If you did not obtain an answer for part (b)(i) use-50.0kJ \(mol^{-1}\), but this is not the correct value.)
▶️Answer/Explanation
Ans:
The accepted value for the enthalpy of combustion of sulfur (\(ΔH_c\)) is \( -297 \, \text{kJ/mol}\),
The experimental value for the enthalpy of combustion of sulfur (\(ΔH_c\)) is \( -31.4 \, \text{kJ/mol}\).
Using the provided values, the percentage error can be calculated as follows:
\[ \text{Percentage error} = \left| \frac{-297 \, \text{kJ/mol} – (-31.4 \, \text{kJ/mol})}{-297 \, \text{kJ/mol}} \right| \times 100\% \]
\[ \text{Percentage error} = \left| \frac{-297 \, \text{kJ/mol} + 31.4 \, \text{kJ/mol}}{-297 \, \text{kJ/mol}} \right| \times 100\% \]
\[ \text{Percentage error} = \left| \frac{-265.6 \, \text{kJ/mol}}{-297 \, \text{kJ/mol}} \right| \times 100\% \]
\[ \text{Percentage error} = \left| 0.894 \right| \times 100\% \]
\[ \text{Percentage error} = 89.4\% \]