Question

Given that ΔG∘ = −13.6 kJ/mol, calculate ΔG at 25∘C for the following sets of conditions.

1) 30 atm NH3, 30 atm CO2, 4.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures.

2) 8.0×10−2 atm NH3, 8.0×10−2 atm CO2, 1.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures.

Is the reaction spontaneous for the conditions in Part A and/or Part B?

A) Is the reaction spontaneous for the conditions in Part A and/or Part B?

B) spontaneous for the conditions in both parts.

C) spontaneous for the conditions in Part but nonspontaneous for the conditions in Part B. spontaneous for the conditions in Part B but nonspontaneous for the conditions in Part A

D) nonspontaneous for the conditions in both parts.

Answer 1

Sol.

As Reaction :

2NH3(g) + CO2(g) ----> NH2CONH2(aq) + H2O(l)

Part A :  

Reaction Qoutient = Q

= [NH2CONH2] / ( (PNH3)² × PCO2 )

= 4.0 / ( (30)² × 30 )  

= 0.0001481

Also, Gas constant = R = 0.008314 KJ / K mol

Temperature = T = 25°C =  298 K   

Standard free energy of reaction = deltaG° = - 13.6 KJ / mol  

So , free energy of reaction = deltaG

= deltaG° + R T ln(Q)

= - 13.6 + 0.008314 × 298 × ln(0.0001481)

= - 35.446 KJ / mol

Part B :

Reaction Qoutient = Q

= [NH2CONH2] / ( (PNH3)² × PCO2 )

= 1.0 / ( (8.0 × 10^-2)² × 8.0 × 10^-2 )  

= 15625

Also, Gas constant = R = 0.008314 KJ / K mol

Temperature = T = 25°C =  298 K   

Standard free energy of reaction = deltaG° = - 13.6 KJ / mol  

So , free energy of reaction = deltaG

= deltaG° + R T ln(Q)

= - 13.6 + 0.008314 × 298 × ln(15625)

=   10.325   KJ / mol

As deltaG is negative in part A , so , reaction will be spontaneous for conditions in part A  

And , deltaG is positive in part B , so , reaction will be non -spontaneous for conditions in part B