SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

CHEMISTRY, M4 EQ-Bank 17

The decomposition of a metal carbonate is represented by the following equation:

\(\ce{MCO3(s) → MO(s) + CO2(g)}\)

The following data was recorded:

\(\Delta H = +130 \, \text{kJ/mol},\ \ \Delta S = +160 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy at 350 K.   (2 marks)
  1. Determine if the reaction is spontaneous at this temperature.   (1 mark)
  1. Discuss how both enthalpy and entropy influence the spontaneity of this reaction and predict the temperature range in which the reaction will be spontaneous.   (4 marks)
Show Answers Only

a.    \(\Delta G = +74 \, \text{kJ/mol}\)

b.    The reaction is non-spontaneous at 350 K.

c.   Enthalpy and entropy influence the spontaneity:

  • To be spontaneous, \(\Delta G\) must be negative, where  \(\Delta G = \Delta H- T\Delta S\).
  • The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.
  • However, the positive entropy means the disorder of the system increases, which favours spontaneity.
  • At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.
  • To find the temperature where the reaction becomes spontaneous, set \(\Delta G = 0\):
  •    \(0=\Delta H-T\Delta S\)
  •    \(T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}\)
  • Thus, the reaction will be spontaneous above 812.5 K.
Show Worked Solution

a.    \(\Delta G = \Delta H- T\Delta S\)

Convert \(\Delta S\) to \(\text{kJ/mol K}\)

\(\Delta S = 0.160 \, \text{kJ/mol K}\)

\(\Delta G = 130-(350 \times 0.160) = 130-56 = +74 \, \text{kJ/mol}\)
 

b.    Since \(\Delta G > 0\), the reaction is non-spontaneous at 350 K.
 

c.   Enthalpy and entropy influence the spontaneity:

  • To be spontaneous, \(\Delta G\) must be negative, where  \(\Delta G = \Delta H- T\Delta S\).
  • The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.
  • However, the positive entropy means the disorder of the system increases, which favours spontaneity.
  • At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.
  • To find the temperature where the reaction becomes spontaneous, set \(\Delta G = 0\):
  •    \(0=\Delta H-T\Delta S\)
  •    \(T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}\)
  • Thus, the reaction will be spontaneous above 812.5 K.

CHEMISTRY, M4 EQ-Bank 16

Given the following values for a reaction:

\(\Delta H = +150 \, \text{kJ/mol}\)  and  \(\Delta S = +250 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy at 298 K.   (2 mark)
  2. Is the reaction spontaneous at this temperature? Justify your answer.   (1 mark)

  3. Predict the temperature at which the reaction would become spontaneous.   (2 marks)
Show Answers Only

a.    \(\Delta G = +75.5 \, \text{kJ/mol}\)
 
b.    The reaction is non-spontaneous at 298 K.
 
c.    The reaction becomes spontaneous above 600 K.

Show Worked Solution

a.    \(\Delta G = \Delta H-T\Delta S\):

Convert \(\Delta S\) to \(\text{kJ/mol K}\):

\(\Delta S = 0.250 \, \text{kJ/mol K}\)

\(\Delta G = 150-(298 \times 0.250) = 150-74.5 = +75.5 \, \text{kJ/mol}\)
 

b.    \(\Delta G > 0\) at 298 K \(\Rightarrow\) reaction is non-spontaneous at 298 K.
 

c.    Find the temperature where the reaction becomes spontaneous:

  • Solve for \(T\) when \(\Delta G = 0\):
  •    \(0= \Delta H-T\Delta S\)
  •    \(T = \dfrac{\Delta H}{\Delta S} = \dfrac{150}{0.250} = 600 \, \text{K}\)
  • As the temperature rises, \(T\Delta S\) increases, therefore \(\Delta G\) decreases, increasing the spontaneity of the reaction.
  • The reaction becomes spontaneous above 600 K.

CHEMISTRY, M4 EQ-Bank 15

A wide range of chemical reactions take place in the combustion chamber of car engines. One such reaction occurs when carbon monoxide is converted to carbon dioxide:

\(\ce{2CO(g) + O2(g) \rightarrow 2CO2(g)}\)

The following values were obtained under standard conditions:

\(\Delta H = +283 \, \text{kJ/mol}, \quad \Delta S = +86.6 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy for the above reaction at 298 K.   (2 marks)
  1. Explain the effect of enthalpy and entropy on the spontaneity of this reaction.   (3 marks)
Show Answers Only

a.    \(\Delta G = +257.2 \, \text{kJ/mol}\)

b.    Effect of enthalpy and entropy on spontaneity:

  • Since the entropy value for this reaction is positive, the reaction is considered to be entropy-driven as this will drive towards a negative value for the Gibbs free energy and contribute to the reaction being spontaneous.
  • However, the enthalpy value of this reaction is also positive, which does not contribute towards a negative value for the Gibbs free energy and in turn contributes towards making this reaction non-spontaneous.
  • Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.
Show Worked Solution

a.    Using the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

Convert \(\Delta S\) to \(\text{kJ/mol K}\):

\(\Delta S = +0.0866 \, \text{kJ/mol K}\).

Now, substitute the values into the equation:

\(\Delta G = 283 \, \text{kJ/mol}-(298 \, \text{K} \times 0.0866 \, \text{kJ/mol K}) = 283-25.8 = +257.2 \, \text{kJ/mol}\)

  • Since \(\Delta G > 0\), the reaction is non-spontaneous.

b.    Effect of enthalpy and entropy on spontaneity:

  • Since the entropy value for this reaction is positive, the reaction is considered to be entropy-driven as this will drive towards a negative value for the Gibbs free energy and contribute to the reaction being spontaneous.
  • However, the enthalpy value of this reaction is also positive, which does not contribute towards a negative value for the Gibbs free energy and in turn contributes towards making this reaction non-spontaneous.
  • Since the magnitude of the enthalpy value is greater, the entropy drive is unable to overcome this and results in a non-spontaneous reaction overall.

CHEMISTRY, M4 EQ-Bank 14

The combustion of propane is represented by the following equation:

\(\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)}\)

\(\Delta H = -2220\ \text{kJ mol}^{-1}\),  \(\Delta S = -269\ \text{J mol}^{-1}\text{K}^{-1}\)

  1. Justify why the entropy change for this reaction is negative based on the chemical equation.   (2 marks)
  1. Using the Gibbs free energy equation, determine if the combustion of propane is spontaneous at 298 K. Show your working.  (2 marks)
Show Answers Only

a.    Entropy change in the combustion of propane:

  • The reactants include 1 molecule of propane gas and 5 molecules of oxygen gas, making a total of 6 gas molecules.
  • The products, however, consist of 3 molecules of carbon dioxide gas and 4 molecules of liquid water.
  • Since gases have higher entropy than liquids, the total number of gas molecules decreases from 6 to 3, which results in a decrease in disorder.
  • This explains why the entropy change, \(\Delta S\), is negative. 

b.    Using the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

Convert \(\Delta S\) from \(\text{J mol}^{-1}\text{K}^{-1}\) to \(\text{kJ mol}^{-1}\text{K}^{-1}\):

\(\Delta S = -0.269\ \text{kJ mol}^{-1}\text{K}^{-1}\).

\(\Delta G\) \(= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})\)  
  \(= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}\)  
  \(= -2139.84\ \text{kJ mol}^{-1}\)  

 
\(\Rightarrow\) Since \(\Delta G < 0\), the reaction is spontaneous at 298 K.

Show Worked Solution

a.    Entropy change in the combustion of propane:

  • The reactants include 1 molecule of propane gas and 5 molecules of oxygen gas, making a total of 6 gas molecules.
  • The products, however, consist of 3 molecules of carbon dioxide gas and 4 molecules of liquid water.
  • Since gases have higher entropy than liquids, the total number of gas molecules decreases from 6 to 3, which results in a decrease in disorder.
  • This explains why the entropy change, \(\Delta S\), is negative.

b.    Using the Gibbs free energy equation:

\(\Delta G = \Delta H-T \Delta S\)

Convert \(\Delta S\) from \(\text{J mol}^{-1}\text{K}^{-1}\) to \(\text{kJ mol}^{-1}\text{K}^{-1}\):

\(\Delta S = -0.269\ \text{kJ mol}^{-1}\text{K}^{-1}\).

\(\Delta G\) \(= -2220\ \text{kJ mol}^{-1}-(298\ \text{K} \times -0.269\ \text{kJ mol}^{-1}\text{K}^{-1})\)  
  \(= -2220\ \text{kJ mol}^{-1} + 80.162\ \text{kJ mol}^{-1}\)  
  \(= -2139.84\ \text{kJ mol}^{-1}\)  
     
  • Since \(\Delta G < 0\), the reaction is spontaneous at 298 K.

CHEMISTRY, M4 EQ-Bank 11

Consider the reaction below at 298 K:

\(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\)

The standard enthalpies of formation and standard entropies are as follows:

\(\begin{aligned}
\ce{N2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 191.5 \, \text{J/mol·K} \\
\ce{H2(g)}: & \quad \Delta H_f^\circ = 0 \, \text{kJmol}^{-1}, & S^\circ = 130.7 \, \text{J/mol·K} \\
\ce{NH3(g)}: & \quad \Delta H_f^\circ = -45.9 \, \text{kJmol}^{-1}, & S^\circ = 192.3 \, \text{J/mol·K} \\
\end{aligned}\)

Calculate the Gibbs free energy (\( \Delta G \)) for the reaction and determine whether the reaction is spontaneous at 298 K. Show all relevant calculations.   (4 marks)

Show Answers Only

\(\Delta G =-32.7\ \text{kJ/mol} \), indicating the reaction is spontaneous at 298 K.

Show Worked Solution

To calculate the Gibbs free energy (\( \Delta G \)) for the reaction, we use the following equation:

   \( \Delta G = \Delta H-T \Delta S \)

First, calculate the change in enthalpy (\( \Delta H \)) for the reaction:

   \( \Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ (\text{products}) -\ \sum \Delta H_f^\circ (\text{reactants})\)

   \( \Delta H_{\text{reaction}}^\circ = [2 \times (-45.9)]-[1 \times (0) + 3 \times (0)] = -91.8 \, \text{kJ/mol}\)
 

Calculate the change in entropy (\( \Delta S \)) for the reaction:

\( \Delta S_{\text{reaction}}^\circ\) \(= \sum S^\circ (\text{products})-\sum S^\circ (\text{reactants}) \)  
  \(= (2 \times (192.3))-(1 \times (191.5) + 3 \times (130.7))\)  
  \(= 384.6-(191.5 + 392.1)\)  
  \(= -199.0\ \text{J/mol·K}\)  
  \(=-0.199\ \text{kJ/mol·K}\)  

 
Calculate \( \Delta G \):

\(\Delta G= \Delta H-T \Delta S= -91.8-(298 \times -0.199)= -32.5\ \text{kJ/mol}\)

  • The reaction is spontaneous at 298 K.

CHEMISTRY, M4 EQ-Bank 5

The following reaction represents the conversion of diamond to graphite:

\(\ce{2C_{diamond} \rightarrow 2C_{graphite}}\)

    • \(\ce{\Delta $H$_{f}\ C_{diamond} = 1.9 kJ mol^{-1}}\)
    • \(\ce{\Delta $S$_{f}\ C_{diamond} = 2.38 J mol^{-1} K^{-1}}\)
    • \(\ce{\Delta $S$_{f}\ C_{graphite} = 5.74 J mol^{-1} K^{-1}}\)
  1. Determine \(\Delta G\) at 298K and state whether the reaction is spontaneous or not.   (3 marks)

  2. What does \(\Delta G\) indicate about the rate of reaction?   (1 mark)

Show Answers Only

a.    \(\Delta G = -5.8025\ \text{kJ}\)

  • Reaction is spontaneous.

b.   Rate of reaction and \(\Delta G\): 

  • Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year
Show Worked Solution

a.  Standard enthalpy and entropy of elements in their natural state is 0.

\(\Delta H\) \(= \Sigma H_{\text{products}}-\Sigma H_{\text{reactants}}\)  
  \(= (2 \times 0)-(2 \times 1.9)\)  
  \(=-3.8\ \text{kJ mol}^{-1} \)  

 

\(\Delta S\) \(=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}\)  
  \(= (2 \times 5.74)-(2 \times 2.38)\)  
  \(= 11.48-4.76 \)  
  \(=6.72\ \text{J mol}^{-1}\ \text{K}^{-1}\)  

 

\(\Delta G\) \(= \Delta H-T\Delta S\)  
  \(= -3.8-(298 \times 0.00672)\)  
  \(= -5.8025\ \text{kJ}\)  
     
  •  The reaction is spontaneous as \(\Delta G < 0\).
     

b.   Rate of reaction and \(\Delta G\): 

  • Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year

CHEMISTRY, M4 EQ-Bank 1

Find \(\Delta G \) at 298.15K for the following reaction:

\(\ce{2CO + O2 \rightarrow 2CO2}\)

Given that  \(\Delta H =-128.3\ \text{kJ},\ \ \Delta S = -159.5\ \text{J K}^{-1} \)  (2 marks)

Show Answers Only

\(-80.75\ \text{kJ}\)

Show Worked Solution

\(\Delta S = -159.5\ \text{J K}^{-1} = -0.1595\ \text{kJ K}^{-1}\)

\(\text{Using the Gibbs free energy equation:}\)

\(\Delta G\) \(= \Delta H-T \Delta S\)  
  \(=-128.3-(298.15 \times -0.1595) \)  
  \(=-128.3 + 47.554\)  
  \(=-80.75\ \text{kJ}\)