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Using Hess’s Law, if the enthalpy changes for the following reactions are known:
\(1.\ \ce{C(s) + O2(g) -> CO2(g)}\ \ \ \ \ \ \ \ \ \ \ \Delta H_1 = -393\ \text{kJ mol}^{-1}\)
\(2.\ \ce{CO(g) + 1/2 O2(g) -> CO2(g)}\ \ \ \ \Delta H_2 = -283\ \text{kJ mol}^{-1}\)
Determine the enthalpy change for the reaction:
\(\ce{C(s) + 1/2 O2(g) -> CO(g)}\)
\(B\)
→ Reverse equation 2, making \(\ce{Co(g)}\) a product and it changes the sign for \(\Delta H_2\).
Equation 2*: \(\ce{CO2(g) -> C2(g) + 1/2 O2(g)} \qquad -\Delta H_2 = +283\ \text{kJ mol}^{-1}\)
→ Combine equation 2* with equation 1 which cancels out \(\ce{CO2(g)}\) and adds the enthalpy values together.
→ Thus, the enthalpy for the final reaction \(=-393 + 283= -110\ \text{kJ mol}^{-1}\)
\(\Rightarrow B\)
Which statement best describes Hess's Law?
\(C\)
→ Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final states are the same.
→ This means that whether a reaction happens in one step or multiple steps, the overall enthalpy change will be the same if the reactants and products are the same.
\(\Rightarrow C\)
Determine the enthalpy change \((\Delta H_4)\) for the following reaction
\(\ce{CH4(g) -> C(s) + 2H2(g)}\)
Using the given chemical reactions and their associated enthalpy changes
| \(1.\) | \(\ce{2CH4(g) + 4O2(g) -> 2CO2(g) +4H2O(l)}\) | \(\Delta H_1=-1780\ \text{kJ mol}^{-1}\) |
| \(2.\) | \(\ce{C(s) + O2 (g) -> CO2(g)}\) | \(\Delta H_2=-393\ \text{kJ mol}^{-1}\) |
| \(3.\) | \(\ce{2H2(g) + O2(g) -> 2H2O(l)}\) | \(\Delta H_3=-572\ \text{kJ mol}^{-1}\) |
\(B\)
\(\frac{1}{2} \times\ \text{Equation 1:}\)
\(\ce{CH4(g) + 2O2(g) -> CO2(g) +2H2O(l)}\) \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJ mol}^{-1}\)
\(\text{Reverse Equation 2 and Equation 3:}\)
\(\ce{CO2(g) -> C(s) + O2(g)}\) \(-\Delta H_2=393\ \text{kJ mol}^{-1}\)
\(\ce{2H2O(l) -> 2H2(g) + O2(g)}\) \(-\Delta H_3=572\ \text{kJ mol}^{-1}\)
\(\text{Add the equations and their}\ \Delta H\ \text{values and cancel any reactants and products:}\)
| \( \ce{CH4(g)} + \cancel{ \ce{2O2(g)}}\) | \( \rightarrow \cancel{\ce{CO2(g)}} + \cancel{ \ce{H2O(l)}}\) | \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJmol}^{-1}\) |
| \(\ce{2C(s)} +\cancel{\ce{2O2(g)}}\) | \(\rightarrow \cancel{ \ce{2CO2(g)}}\) | \(-\Delta H_2=393\ \text{kJ mol}^{-1}\) |
| \(\ce{H2(g)} + \cancel{\ce{\frac{1}{2}O2(g)}}\) | \(\rightarrow \cancel{\ce{H2O(l)}}\) | \(-\Delta H_3=572\ \text{kJ mol}^{-1}\) |
| \(\ce{CH4(g)}\) | \(\ce{\rightarrow C(s) + 2H2(g)}\) | \(\Delta H_4=75\ \text{kJ mol}^{-1}\) |
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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.
\(\ce{C2H6(g) + 2.5 O2(g) → 2 CO(g) + 3 H2O(l)}\)
\(\Delta H = \Delta H_f \text{ products}-\Delta H_f \text{ reactants}\)
\(\Delta H = [2(-110.5) + 3(-285.8)]-[(-84.7)] = -1177.6 \, \text{kJ/mol}\)
Three equations and their \(\Delta H\) values are shown.
| Equation 1: | \(\ce{C(s) + O2(g) → CO2(g)}\) | \(\Delta H_1 = -393.5 \, \text{kJ mol}^{-1}\) |
| Equation 2: | \(\ce{H2(g) + 1/2 O2(g) → H2O(l)}\) | \(\Delta H_2 = -285.8 \, \text{kJ mol}^{-1}\) |
| Equation 3: | \(\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)}\) | \(\Delta H_3 = -890.4 \, \text{kJ mol}^{-1}\) |
Using this information, what is the \(\Delta H\) for the following reaction?
\(\ce{C(s) + 2 H2(g) → CH4(g)}\)
\(B\)
Reverse Equation 3 to express \(\ce{CH4(g)}\) as a product:
Equation 3*: \(\ce{CO2(g) + 2 H2O(l) → CH4(g) + 2 O2(g)} \quad -\Delta H_1 = +890.4 \, \text{kJ mol}^{-1}\)
Double the coefficients of Equation 2 which also doubles \(\Delta H_2\).
Equation 2*: \(\ce{2H2(g) + O2(g) → 2H2O(l)} \qquad 2\Delta H_2 = -571.6 \, \text{kJ mol}^{-1}\)
Combine the corresponding enthalpies of Equation 1, 2* and 3* to find the total \(\Delta H\):
\(\Delta H = +890.4-393.5-571.6 = -74.9 \, \text{kJ mol}^{-1}\)
\(\Rightarrow B\)
Enthalpy changes for the melting of iodine, \(\ce{I2}\), and for the sublimation of iodine are provided below.
\(\ce{I2(s) \rightarrow I2(l) \quad \quad \Delta H = +16 kJ mol^{-1}}\)
\(\ce{I2(s) \rightarrow I2(g) \quad \quad \Delta H = +62 kJ mol^{-1}}\)
Determine the enthalpy change for the vaporisation of iodine that is represented by the equation \(\ce{I2(l) \rightarrow I2(g)}\). (2 marks)
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\(+46\ \text{kJ mol}^{-1}\)
\(\ce{I2(l) \rightarrow I2(s)}, \quad \Delta H = -16\ \text{kJ mol}^{-1}\)
\(\ce{I2(s) \rightarrow I2(g)}, \quad \Delta H = +62\ \text{kJ mol}^{-1}\)
\(\ce{I2(l) \rightarrow I2(g)}, \quad \Delta H = -16 +62 = +46\ \text{kJ mol}^{-1}\)
Nitrogen dioxide decomposes as follows.
\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g)}\ \quad \quad \Delta H = -66 \text{ kJ mol}^{-1}\)
The enthalpy change for the reaction represented by the equation \(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g)}\) is
\(C\)
\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g),}\ \quad \Delta H = -66 \text{ kJ mol}^{-1}\)
\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g),}\ \quad \Delta H = +66 \text{ kJ mol}^{-1}\)
\(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g),}\ \quad \Delta H = +33 \text{ kJ mol}^{-1}\)
\(\Rightarrow C\)
The combustion of hexane takes place according to the equation
\(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \quad \Delta H = -4158\ \text{kJ mol}^{-1}\)
Consider the following reaction.
\(\ce{ 12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\)
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a. \(\Delta H = +8316\ \text{kJ mol}^{-1}\)
b. Endothermic
a. \(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \Delta H = -4158\ \text{kJ mol}^{-1}\)
→ Reverse equation:
\(\ce{6CO2(g) + 7H2O(g)\rightarrow C6H14(g) + \dfrac{19}{2}O2(g)}\) \(\quad \Delta H = +4158\ \text{kJ mol}^{-1}\)
→ Double equation:
\(\ce{12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\) \(\quad \Delta H = 2 \times +4158\ \text{kJ mol}^{-1}\)
→ \(\Delta H = +8316\ \text{kJ mol}^{-1}\)
b. Endothermic
Consider the following equations.
| \(\ce{\frac{1}{2}N2(g) + O2(g) → NO2(g)}\) | \(\quad\Delta H = +30 \text{kJ mol}^{-1}\) |
| \(\ce{N2(g) + 2O2(g) → N2O4(g)}\) | \(\quad \Delta H = +10 \text{kJ mol}^{-1}\) |
Calculate the enthalpy change for the reaction \(\ce{N2O4(g)\rightarrow 2NO2(g)}\) (2 marks)
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\(\Delta H = + 50\ \text{kJ mol}^{-1}\)
→ Double 1st equation:
\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g)\ \ \ \ \Delta H = +60\ \text{kJ mol}^{-1}}\)
→ Reverse 2nd equation:
\(\ce{N2O4(g) \rightarrow N2(g) + 2O2(g)\ \ \ \ \Delta H = -10\ \text{kJ mol}^{-1}}\)
→ Combining both equations:
\(\ce{N2O4(g)\rightarrow 2NO2(g)\ \ \ \ \Delta H = +50\ \text{kJ mol}^{-1}}\)
The enthalpies of reaction of a number of chemical reactions are as follows:
Reaction 1: \(\ce{CH3CH2OH(l) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(l)}\) \(\Delta H_1=-1360\ \text{kJ mol}^{-1}\)
Reaction 2: \(\ce{CH3COOH(l) + 2O2(g) \rightarrow 2CO2(g) + 2H2O(l)}\) \(\Delta H_2=-876\ \text{kJ mol}^{-1}\)
Reaction 3: \(\ce{H2(g) + \frac{1}{2}O2(g) \rightarrow H2O(l)}\) \(\Delta H_3=-285.8\ \text{kJ mol}^{-1}\)
Calculate the enthalpy change for the reaction below using the enthalpies of reaction above.
\(\ce{CH3COOH(l) + H2(g) \rightarrow CH3CH2OH(l) + \frac{1}{2}O2(g)}\) \(\Delta H_4=?\) (3 marks)
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\(\Delta H_4=+198.2\ \text{kJ mol}^{-1}\)
→ Reverse equation 1 to make \(\ce{CH3CH2OH(l)}\) a product (changes the sign of the enthalpy).
\(\ce{2CO2(g) + 3H2O(l) \rightarrow CH3CH2OH(l) + 3O2(g)}\) \(-\Delta H_1=+1360\ \text{kJ mol}^{-1}\)
→ Add the equations and their \(\Delta H\) values and cancel any reactants and products.
\begin{array} {r l l}
\rule{0pt}{2.5ex} \cancel{ \ce{2CO2(g)}} + \cancel{ \ce{3H2O(l)}} \rule[-1ex]{0pt}{0pt} & \rightarrow \ce{CH3CH2OH(g)} + \cancelto{\ce{\frac{1}{2}O2(g)}}{\ce{3O2(g)}} & -\Delta H_1=+1360\ \text{kJ mol}^{-1}\\
\rule{0pt}{2.5ex} \ce{CH3COOH(l)} +\cancel{\ce{2O2(g)}} \rule[-1ex]{0pt}{0pt} & \rightarrow \cancel{ \ce{2CO2(g)}} + \cancel{\ce{2H2O(l)}} & 2\Delta H_2= -876\ \text{kJ mol}^{-1} \\
\rule{0pt}{2.5ex} \ce{H2(g)} + \cancel{\ce{\frac{1}{2}O2(g)}} \rule[-1ex]{0pt}{0pt} & \rightarrow \cancel{\ce{H2O(l)}} & \Delta H_3=-285.8\ \text{kJ mol}^{-1} \\
\rule{0pt}{2.5ex}\ce{CH3COOH(l) + H2(g)} \rule[-1ex]{0pt}{0pt} & \ce{\rightarrow CH3CH2OH(l) + \frac{1}{2}O2(g)} & \Delta H_4=198.2\ \text{kJ mol}^{-1} \\
\end{array}
The enthalpies of reaction of a number of chemical reactions are as follows:
Reaction 1: \(\ce{C2H2(g) + \frac{5}{2}O2(g) \rightarrow 2CO2(g) + H2O(l)}\) \(\Delta H_1=-1299.5\ \text{kJ mol}^{-1}\)
Reaction 2: \(\ce{C(s) + O2(g) \rightarrow CO2(g)}\) \(\Delta H_2=-393.5\ \text{kJ mol}^{-1}\)
Reaction 3: \(\ce{H2(g) + \frac{1}{2}O2(g) \rightarrow H2O(l)}\) \(\Delta H_3=-285.8\ \text{kJ mol}^{-1}\)
Calculate the enthalpy change for the reaction below, stating whether the reaction is exothermic or endothermic:
\(\ce{2C(s) + H2(g) \rightarrow C2H2(g)}\) \(\Delta H_4=?\) (4 marks)
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\(\Delta H_4=+226.7\ \text{kJ mol}^{-1}\)
→ Reverse equation 1 to make \(\ce{C2H2(g)}\) a product (changes the sign of the enthalpy).
\(\ce{2CO2(g) + H2O(l) \rightarrow C2H2(g) + \frac{5}{2}O2(g)}\) \(-\Delta H_1=+1299.5\ \text{kJ mol}^{-1}\)
→ Add a multiplier to Reaction 2 as this will allow the \(\ce{CO2(g)}\) to cancel when applying Hess’s Law.
\(\ce{2C(s) +2O2(g) \rightarrow 2CO2(g)}\) \(2\Delta H_2=2 \times -393.5= -787\ \text{kJ mol}^{-1}\)
→ Add the equations and their \(\Delta H\) values and cancel any reactants and products.
| \(\cancel{ \ce{2CO2(g)}} + \cancel{ \ce{H2O(l)}}\) | \( \rightarrow \ce{C2H2(g)} + \cancel{ \ce{\frac{5}{2}O2(g)}}\) | \(-\Delta H_1=+1299.5\ \text{kJ mol}^{-1}\) |
| \(\ce{2C(s)} +\cancel{\ce{2O2(g)}}\) | \(\rightarrow \cancel{ \ce{2CO2(g)}}\) | \(2\Delta H_2= -787\ \text{kJ mol}^{-1}\) |
| \(\ce{H2(g)} + \cancel{\ce{\frac{1}{2}O2(g)}}\) | \(\rightarrow \cancel{\ce{H2O(l)}}\) | \(\Delta H_3=-285.8\ \text{kJ mol}^{-1}\) |
| \(\ce{2C(s) + H2(g)}\) | \(\ce{\rightarrow C2H2(g)}\) | \(\Delta H_4=+226.7\ \text{kJ mol}^{-1}\) |
→ The reaction is endothermic.
Define Hess's Law with reference to reaction pathways and changes in enthalpy. (2 marks)
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→ The enthalpy change in a chemical reaction between the reactants and products is constant, regardless of the number of pathways or chemical reactions undertaken to get from the reactants to the products.
→ The enthalpy change in a chemical reaction between the reactants and products is constant, regardless of the number of pathways or chemical reactions undertaken to get from the reactants to the products.