Question

In Class Exercise - The Gibbs Free Energy Change, AG 1) Determining the Standard Gibbs Free Energy Change (AGⓇ) for a Chemical Reaction 2) Using AGº to Determine Spontaneity Name: Date: Lab section: Show your work when there are calculations, write units, and use correct significant figures. Consider the following reaction (balanced as written) and thermodynamic data from tables in your book: CO(NH2)2(aq) + H2O(1) ► CO2(g) + 2NH3(g) Substance CO(NH3)2(aq) H00 AH(kJ/mol) 1-391.2 -285.9 -3935 -46.19 S'J/mol K) 173.8 69.96 213.6 192.5 CO(g) NH 1) What is the standard Gibbs free energy change for this reaction at 25°C? 2) This reaction is nonspontaneous at 25.0 °C but becomes spontaneous at a higher temperature. Calculate this temperature and report it in 'C.

Answer 1

$\Delta G^{o} = \Delta H^{o} - T\Delta S^{o}$

where $\Delta$ G^o is standard Gibbs free energy change,  $\Delta$H^o is standard change in enthalpy ,  $\Delta$S^o is standard change in entropy.

Given temperature = 273.15 + 25 = 298.15 K

We need to calculate the overall change in enthalpy and entropy of the reaction to find $\Delta$ G^o.

$\Delta$H^o reaction = $\Delta$ H^oproduct - $\Delta$ H^o reactant = ( -393.5 - 46 ) - ( 391.2 - 285.9 ) KJ / mol = 237.6 KJ/mol.

$\Delta$S^o reaction = $\Delta$ S^o product - $\Delta$ S^o reactant = ( 213.6 + 192.5 ) - ( 69.96 + 173.8 ) J / mol.K = 162.34 J / mol.K

$\Delta$G^o =( 237.6 x 1000 ) J/mol - ( 298.15 K x 162.34 J / mol K ) = 189198.32 J/mol = 189.2 KJ / mol

$\Delta$G^o is positive, hence the reaction is non-spontaneous.

2) For the reaction to be spontaneous $\Delta$ G^o needs to be negative. If we equate $\Delta$ G^o to zero we can find out the temperature for which the reaction will be spontaneous.

$\Delta$G^o = $\Delta$ H^o - T$\Delta$S^o = 0

( 237.6 x 1000 ) J / mol - ( T x 162.34 J/mol K ) = 0

solving gives T = ( 237.6 x 1000 J/mol) / ( 162.34 J/ mol x K ) = 1463.59 K = 1190.44 ^oC