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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)

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

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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

Filed Under: Entropy and Gibbs Free Energy Tagged With: Band 4, Band 5, smc-4268-50-Determining Spontaneity, smc-4268-80-Calculating G, smc-4268-80-Calculating S

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