Gibbs Free Energy/Enthelpy/Entropy Question

[vancity94]

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.

 

[Ki-nana18]

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.

 

[vancity94]

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.

 

[Borek]

Significant digits perhaps?

 

[DeeNos]

Can you help me with the correct approach to this problem?Your approach is correct, but you made a mistake in calculating ΔS. It should be -147.65 J K^-1 mol^-1, not kJ K^-1 mol^-1. Also, when plugging in the values for ΔH and ΔS into the ΔG equation, it should be in units of kJ, not kJ mol^-1. So the correct calculation would be:

ΔH = (64.77) + 0 - 2(71.67) = -78.57 kJ
ΔS = (33.15) + (-99.6) - 2(40.6) = -147.65 J K^-1
ΔG = ΔH - TΔS
= -78.57 - 533(-0.14765)
= -78.57 + 78.57 = 0 kJ

So the ΔG at 533 K for this reaction is 0 kJ, indicating that the reaction is at equilibrium at this temperature.