Question

1 Find out the value of equilibrium constant for the following reaction at 298 K
2NH3(g)+CO2(g)=NH2CONH2(aq)+H2O(l)
Standard Gibbs energy change at the given temperature is -13.6kJ per mole

Answer 1

Answer:

1. Predict whether the entropy change will be positive or negative for the following:

a. H2O (g)  H2O (l) S__-__

b. C6H12O6(s)  2C2H5OH(l) + 2CO2(g) S__+_

c. 2NH3(g) + CO2(g)  H2O(l) + NH2CONH2(aq) S__-__

d. NaCl(s)  NaCl(aq) S__+__

e. Cu(s) (100oC)  Cu(s) (25oC) S__-__

f. 2NH3(g) N2(g) + 3H2(g) S__+__

2. Calculate ∆H°rxn for C6H12O6(s)  2C2H5OH(l) + 2CO2(g) using ∆H°f values.

∆H°rxn =[(2 mol)(-277.7kJ/mol)+(2 mol)(-393.5kJ/mol)]-(1mol)(-1273.02 kJ/mol) =-69.38kJ

3. Calculate ∆S°rxn and ∆S°surr.

∆S°rxn =[(2 mol)(160.7 J/molK)+(2 mol)(213.6 J/molK)]-(1mol)(212.1 J/molK)=536.5 J/K

232.7 J/K

4. Based on your values for ∆S°rxn and ∆S°surr, is the reaction spontaneous under standard

thermodynamic conditions?

Yes, ∆S°rxn/sys + ∆S°surr = 536.5 J/K + 232.7 J/K = 769.2 J/K. Since this value is >0, it is

spontaneous.

5. Is this reaction always spontaneous? If not, determine at what temperatures it changes from

spontaneous to non-spontaneous.

With a -∆H°, and +∆S°, this reaction will be spontaneous under all conditions.

6. For the following reaction:

2Mg(s) + O2(g)  MgO(s) , ΔHo

rxn = -1202 kJ/mol; ΔSo

rxn = -217 J/mol∙K

 Calculate ∆G° using ∆G=∆H-T∆S

∆G=-1202 kJ/mol – (298.15K)(-0.217 kJ/mol K) = -1137 kJ/mol

∆Hf° (k

Explanation: