SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

CHEMISTRY, M4 EQ-Bank 10 MC

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:

\(\ce{C(s) + 1/2 O2(g) -> CO(g)}\)
 

  1. \(+110\ \text{kJ mol}^{-1}\)
  2. \(-110\ \text{kJ mol}^{-1}\)
  3. \(-676\ \text{kJ mol}^{-1}\)
  4. \(+676\ \text{kJ mol}^{-1}\)
Show Answer Only

\(B\)

Show Worked Solution
  • 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 \(=-393 + 283= -110\ \text{kJ mol}^{-1}\)

\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 9 MC

Which statement best describes Hess's Law?

  1. The total enthalpy change for a reaction depends on the pathway taken to reach the final products.
  2. The total enthalpy change for a reaction is zero if the reaction is reversible.
  3. The total enthalpy change for a reaction is independent of the pathway taken, as long as the initial and final states are the same.
  4. The enthalpy change for a reaction is the same regardless of whether it occurs in one step or multiple steps, provided heat is lost to the surroundings.
Show Answer Only

\(C\)

Show Worked Solution
  • 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.

\(\Rightarrow C\)

CHEMISTRY, M4 EQ-Bank 5 MC

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

\(1.\)   \(\ce{2CH4(g) + 4O2(g) -> 2CO2(g) +4H2O(l)}\)    \(\Delta H_1=-1780\ \text{kJ mol}^{-1}\)
\(2.\)   \(\ce{C(s) + O2 (g) -> CO2(g)}\)    \(\Delta H_2=-393\ \text{kJ mol}^{-1}\)
\(3.\)   \(\ce{2H2(g) + O2(g) -> 2H2O(l)}\)     \(\Delta H_3=-572\ \text{kJ mol}^{-1}\)

 

  1. \(-75\ \text{kJ mol}^{-1}\)
  2. \(75\ \text{kJ mol}^{-1}\)
  3. \(-485\ \text{kJ mol}^{-1}\)
  4. \(485\ \text{kJ mol}^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\frac{1}{2} \times\ \text{Equation 1:}\)

\(\ce{CH4(g) + 2O2(g) -> CO2(g) +2H2O(l)}\)    \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJ mol}^{-1}\)

\(\text{Reverse Equation 2 and Equation 3:}\)

\(\ce{CO2(g) -> C(s) + O2(g)}\)    \(-\Delta H_2=393\ \text{kJ mol}^{-1}\)

\(\ce{2H2O(l) -> 2H2(g) + O2(g)}\)    \(-\Delta H_3=572\ \text{kJ mol}^{-1}\)
 

\(\text{Add the equations and their}\ \Delta H\ \text{values and cancel any reactants and products:}\)

\( \ce{CH4(g)} + \cancel{ \ce{2O2(g)}}\) \( \rightarrow \cancel{\ce{CO2(g)}} + \cancel{ \ce{H2O(l)}}\)    \(\dfrac{1}{2}\Delta H_1=-890\ \text{kJmol}^{-1}\)
\(\ce{2C(s)} +\cancel{\ce{2O2(g)}}\) \(\rightarrow \cancel{ \ce{2CO2(g)}}\)     \(-\Delta H_2=393\ \text{kJ mol}^{-1}\)
\(\ce{H2(g)} + \cancel{\ce{\frac{1}{2}O2(g)}}\) \(\rightarrow \cancel{\ce{H2O(l)}}\)    \(-\Delta H_3=572\ \text{kJ mol}^{-1}\)
\(\ce{CH4(g)}\) \(\ce{\rightarrow C(s) + 2H2(g)}\)    \(\Delta H_4=75\ \text{kJ mol}^{-1}\)
 
\(\Rightarrow B\)

CHEMISTRY, M4 EQ-Bank 2

  1. Identify Hess' Law.   (2 marks)

  2. 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}

  1. Calculate \(\Delta H\) for the combustion of:
    1.  mole of ethane (\(\ce{C2H6}\)) to produce carbon dioxide and water.   (2 marks)

    2. 1 mole of ethane (\(\ce{C2H6}\)) to produce carbon monoxide and water.   (2 marks)

Show Answers Only

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.   \(\Delta H = -1177.6 \, \text{kJ/mol}\)

Show Worked Solution

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 2 MC

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?

\(\ce{C(s) + 2 H2(g) → CH4(g)}\)

  1. \(+74.9 \, \text{kJ mol}^{-1}\)
  2. \(-74.9 \, \text{kJ mol}^{-1}\)
  3. \(+210.3 \, \text{kJ mol}^{-1}\)
  4. \(-210.3 \, \text{kJ mol}^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

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 \(\Delta H\):

\(\Delta H = +890.4-393.5-571.6 = -74.9 \, \text{kJ mol}^{-1}\)

\(\Rightarrow B\)

CHEMISTRY, M4 2012 VCE 4*

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)

Show Answers Only

\(+46\ \text{kJ mol}^{-1}\)

Show Worked Solution

\(\ce{I2(l) \rightarrow I2(s)}, \quad \Delta H = -16\ \text{kJ mol}^{-1}\)

\(\ce{I2(s) \rightarrow I2(g)}, \quad \Delta H = +62\ \text{kJ mol}^{-1}\)

\(\ce{I2(l) \rightarrow I2(g)}, \quad \Delta H = -16 +62 = +46\ \text{kJ mol}^{-1}\)

CHEMISTRY, M4 2012 VCE 5 MC

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

  1. \(-66 \text{ kJ mol} ^{-1}\)
  2. \(-33 \text{ kJ mol} ^{-1}\)
  3. \(+33 \text{ kJ mol} ^{-1}\)
  4. \(+66 \text{ kJ mol} ^{-1}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g),}\ \quad \Delta H = -66 \text{ kJ mol}^{-1}\)

\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g),}\ \quad \Delta H = +66 \text{ kJ mol}^{-1}\)

\(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g),}\ \quad \Delta H = +33 \text{ kJ mol}^{-1}\)

\(\Rightarrow C\)

CHEMISTRY, M4 2016 VCE 17*

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.

\(\ce{ 12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\)

  1. Calculate the value of \(\Delta H\), in kJ mol\(^{-1}\), for this reaction   (2 marks)

  2. Is the reaction exothermic or endothermic?   (1 mark)

Show Answers Only

a.    \(\Delta H = +8316\ \text{kJ mol}^{-1}\)

b.    Endothermic

Show Worked Solution

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

CHEMISTRY, M4 2015 VCE 18*

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)

Show Answers Only

\(\Delta H = + 50\ \text{kJ mol}^{-1}\)

Show Worked Solution
  • 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:
  •    \(\ce{N2O4(g)\rightarrow 2NO2(g)\ \ \ \ \Delta H = +50\ \text{kJ mol}^{-1}}\)

CHEMISTRY, M4 EQ-Bank 35

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)

Show Answers Only

 \(\Delta H_4=+198.2\ \text{kJ mol}^{-1}\)

Show Worked Solution
  • 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}

CHEMISTRY, M4 EQ-Bank 32

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)

Show Answers Only

 \(\Delta H_4=+226.7\ \text{kJ mol}^{-1}\)

Show Worked Solution
  • 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.

CHEMISTRY, M4 EQ-Bank 31

Define Hess's Law with reference to reaction pathways and changes in enthalpy.   (2 marks)

Show Answers Only
  • 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.
Show Worked Solution
  • 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.