Gibbs Free Energy/Enthalpy/Entropy Question

  • Thread starter vancity94
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  • #1
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Homework Statement



Assume that ΔH° and ΔS° are independent of temperature. Calculate ΔG° for the following reaction at 533 K.

2Cu(+)(aq) ----> Cu(s) + Cu(2+)(aq)

Cu(2+)(aq) enthalpy (H) of formation = 64.77 kJ mol^-1; molar entropy (S) = -99.6 J K^-1 mol^-1
Cu(s) enthalpy (H) of formation = 0 kJ/mol; molar entropy (S) = 33.15 J K^-1 mol^-1
Cu(+)aq enthalpy (H) of formation = 71.67 kJ/mol; molar entropy (S) = 40.6 J K^-1 mol^-1

Homework Equations



ΔG = ΔH - TΔS

The Attempt at a Solution



ΔH = (64.77) + 0 - 2(71.67) = -78.57 kJ mol^-1
ΔS = (33.15) + -99.6 -2(40.6) = -.14765 kJ K^-1 mol^-1

ΔG = ΔH - TΔS
= -78.57 - 533(-.14765)
= .12745 = wrong

I know I'm messing up somewhere because the correct answer is negative.
 

Answers and Replies

  • #2
91
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To find the enthalpy of formation and entropy of formation its the summation of products minus the summation of the reactants, each multiplied by their respective coefficients. Double check your entropy calculation.
 
  • #3
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To find the enthalpy of formation and entropy of formation its the summation of products minus the summation of the reactants, each multiplied by their respective coefficients. Double check your entropy calculation.

I'm still getting -147.65 J K^-1 mol^-1, or -.14765 kJ K^-1 mol^-1.
 
  • #4
Borek
Mentor
28,822
3,340
Significant digits perhaps?
 

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