smc-4267-30-Enthalpies of formation

CHEMISTRY, M4 EQ-Bank 4

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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\(-126\ \text{kJ mol}^{-1}\)

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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 \(-126\ \text{kJ mol}^{-1}\).

CHEMISTRY, M4 EQ-Bank 6 MC

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?

\(\Delta H_f\ \ \ce{ O2(g)}\)
\(\Delta H_f\ \ \ce{ CO2(g)}\)
\(\Delta H_f\ \ \ce{ C6H12O6(s)}\)
\(\Delta H_f\ \ \ce{ H2O(l)}\)

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\(A\)

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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.

\(\Rightarrow A\)

CHEMISTRY, M4 EQ-Bank 1

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}

Define what the term 'standard enthalpy of formation' means. (1 mark)

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Use the table data to calculate the standard enthalpy of formation of butanol. (3 marks)

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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. \(-332\ \text{kJ mol}^{-1}\)

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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.	\(\Delta H\)	\(= \Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}\)
	\(-2670\)	\(=(4 \times -393 + (5 \times -286))-(\Delta H_f \text{ butanol} + (6 \times 0))\)
	\(\Delta H_f \text{ butanol}\)	\(=-3002 + 2670\)
		\(=-332\ \text{kJ mol}^{-1}\)

The standard enthalpy of formation of an element is 0.

CHEMISTRY, M4 EQ-Bank 2

Identify Hess' Law. (2 marks)

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Given these heats of formation, \(\Delta H_f\):

\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}

Calculate \(\Delta H\) for the combustion of:

1. mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water. (2 marks)
  
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2. 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. \(\Delta H = -1177.6 \, \text{kJ/mol}\)

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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. 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.
\(\ce{C2H6(g) + 2.5 O2(g) → 2 CO(g) + 3 H2O(l)}\)
\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)
\(\Delta H = [2(-110.5) + 3(-285.8)]-[(-84.7)] = -1177.6 \, \text{kJ/mol}\)

CHEMISTRY, M4 EQ-Bank 36

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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\(\Delta H^{\circ}_{\text{reaction}} = -71.4\ \text{kJmol}^{-1}\)

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\(\Sigma\ \text{Enthalpy (reactants)}\ = 1 \times -1271 = -1271\ \text{kJ}\)

\(\Sigma\ \text{Enthalpy (products)}\ = 2 \times -277.7 + 2 \times -393.5 = -1342.4\ \text{kJ}\)

\(\Delta H^{\circ}_{\text{reaction}}\)	\(= \Delta H^{\circ} (\text{products})-\Delta H^{\circ} (\text{reactants}) \)
	\(=-1342.4-(-1271)\)
	\(= -71.4\ \text{kJ mol}^{-1}\)