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The chemical equation for the combustion of butane \(\ce{(C4H10)}\) is given below:
\(\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)} \qquad \Delta H = -5754\ \text{kJ mol}^{-1}\)
Given that the standard enthalpy of formation of \(\ce{CO2(g)}\) is –393 kJ mol\(^{-1}\) and \(\ce{H2O(g)}\) is –241 kJ mol \(^{-1}\), calculate the standard enthalpy of formation of butane. (3 marks)
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\(-126\ \text{kJ mol}^{-1}\)
→ The enthalpy change for the combustion of 2 moles of butane is -5754 kJ.
| \(\Delta H\) | \(=\Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}\) | |
| \(-5754\) | \(=(8 \times -393 + 10 \times -241)-(2 \times \Delta H_f \text{ (butane)})\) | |
| \(-5754\) | \(=-4334-2 \times \Delta H_f \text{ (butane)}\) | |
| \(\Delta H_f \text{ (butane)}\) | \(=\dfrac{-4334 + 5754}{2}\) | |
| \(=-126\ \text{kJ mol}^{-1}\) |
→ The standard enthalpy of formation of butane is \(-126\ \text{kJ mol}^{-1}\).
The chemical equation for photosynthesis is given below:
\(\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) +6O2(g)}\)
Which of the following does Not effect the value of \(\Delta H\) for this reaction?
\(A\)
→ The \(\Delta H_f\) of \(\ce{O2}\) is \(0\) because oxygen is in its elemental form.
→ The standard enthalpy of formation of any element in its standard state is zero.
→ Hence it will have no effect on the \(\Delta H\) for photosynthesis.
\(\Rightarrow A\)
The chemical equation for the combustion of butanol \(\ce{(C4H9OH(l))}\) is given below
\(\ce{C4H9OH(l) + 6O2(g) -> 4CO2(g) + 5H2O(l)}\) \(\Delta H = -2670\ \text{kJ mol}^{-1}\)
\begin{array} {|c|c|}
\hline \text{Compound} & \Delta H_f \ \text{(kJ mol}^{-1}) \\
\hline \ce{CO2(g)} & -393 \\
\hline \ce{H2O(l)} & -286 \\
\hline \end{array}
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a. Standard enthalpy of formation:
→ The change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K temperature and 100 kPa).
→ The elements must be in their most stable form at these conditions.
b. \(-332\ \text{kJ mol}^{-1}\)
a. Standard enthalpy of formation:
→ The change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K temperature and 100 kPa).
→ The elements must be in their most stable form at these conditions.
| b. | \(\Delta H\) | \(= \Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}\) |
| \(-2670\) | \(=(4 \times -393 + (5 \times -286))-(\Delta H_f \text{ butanol} + (6 \times 0))\) | |
| \(\Delta H_f \text{ butanol}\) | \(=-3002 + 2670\) | |
| \(=-332\ \text{kJ mol}^{-1}\) |
→ The standard enthalpy of formation of an element is 0.
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\begin{array} {|c|c|}
\hline \text{Chemical} & \Delta H_f \ \text{(kJ mol}^{-1}) \\
\hline \ce{C2H6(g)} & -84.7 \\
\hline \ce{CO2(g)} & -393.5 \\
\hline \ce{H2O(l)} & -285.8 \\
\hline \ce{CO(g)} & -115 \\
\hline \end{array}
i. 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water. (2 marks)
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ii. 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon monoxide and water. (2 marks)
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a. Hess’s Law:
→ The total enthalpy change for a reaction is the same, regardless of the route by which the chemical reaction occurs, provided the initial and final conditions are the same.
b.i. \(\Delta H= -1560.8 \, \text{kJ/mol}\)
b.ii. \(\Delta H = -1177.6 \, \text{kJ/mol}\)
a. Hess’s Law:
→ The total enthalpy change for a reaction is the same, regardless of the route by which the chemical reaction occurs, provided the initial and final conditions are the same.
b.i. Enthalpy change:
→ complete combustion of 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water.
\(\ce{C2H6(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)}\)
\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)
\(\Delta H = [2(-393.5) + 3(-285.8)]-[(-84.7)] = -1560.8 \, \text{kJ/mol}\)
b.ii. Enthalpy change:
→ incomplete combustion of 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon monoxide and water.
\(\ce{C2H6(g) + 2.5 O2(g) → 2 CO(g) + 3 H2O(l)}\)
\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)
\(\Delta H = [2(-110.5) + 3(-285.8)]-[(-84.7)] = -1177.6 \, \text{kJ/mol}\)
The enthalpies of formation for a number of chemical reactions are as follows:
\(\ce{C6H12O6(s)}\) \(\Delta H^{\circ}_f = -1271\ \text{kJmol}^{-1}\)
\(\ce{C2H5OH(aq)}\) \(\Delta H^{\circ}_f = -277.7\ \text{kJmol}^{-1}\)
\(\ce{CO2(g)}\) \(\Delta H^{\circ}_f = -393.5\ \text{kJmol}^{-1}\)
Calculate the enthalpy change for the fermentation of glucose (reaction below) using the enthalpies of formation above.
\(\ce{C6H12O6(s) \rightarrow 2C2H5OH(aq) + 2CO2(g)}\) (3 marks)
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\(\Delta H^{\circ}_{\text{reaction}} = -71.4\ \text{kJmol}^{-1}\)
\(\Sigma\ \text{Enthalpy (reactants)}\ = 1 \times -1271 = -1271\ \text{kJ}\)
\(\Sigma\ \text{Enthalpy (products)}\ = 2 \times -277.7 + 2 \times -393.5 = -1342.4\ \text{kJ}\)
| \(\Delta H^{\circ}_{\text{reaction}}\) | \(= \Delta H^{\circ} (\text{products})-\Delta H^{\circ} (\text{reactants}) \) | |
| \(=-1342.4-(-1271)\) | ||
| \(= -71.4\ \text{kJ mol}^{-1}\) |