R = 0.008314 kJ/K.mol; 1 Faraday (F) = 96, 500 C/mol e^-; 1 J = 1C times 1V. Hyd

Question

R = 0.008314 kJ/K.mol; 1 Faraday (F) = 96, 500 C/mol e^-; 1 J = 1C times 1V. Hydrazine, H_2NNH_2, is used as a rocket fuel. Use the data below to determine if hydrazine be prepared from the following reaction under standard conditions:: N_2(g) + 2 H_2(g) rightarrow H_2NNH_2(l) Calculate Delta S_total for the following reaction, then tell whether it is spontaneous under standard conditions or not. CaCO_3(s) rightarrow CaO(s) + CO_2(g) The entropy change for a certain non-spontaneous reaction at 50 degree C is 104 J/K. Is the reaction endothermic or exothermic? Explain your answer. What is the minimum value of Delta H degree (in kJ) for the reaction? Urea, NH_2CONH_2, can be prepared by the following reaction: 2 NH_3(g)+ CO_2(g) rightarrow NH_2CONH_2 (aq) Given that Delta G degree = -13.6 kJ, calculate Delta G at 25 degree C for the following set of conditions: 10 atm NH_3, 10 atm CO_2, 1.0 M NH_2CONH_2; and 0.10 atm NH_3, 0.10 atm CO_2, 1.0 M NH_2CONH_2 Would get you more product or less product for each condition?

Explanation / Answer

Standard heat of reaction = 50.63 Kj/mole , deltaH= 50.63 Kj/mole

Entropy change =1* entropy of NH2NH2- ( entropy of N2+2* entropy of H2)

= 121.1-(192+2*130.7) =-332.3 J/K

Gibbs Free energy change = deltaH- T*deltas= 50.63*1000+332.3*298 =149655.4 KJ

Since Gibbs free energy change is +ve, the reaction is not spontaneous and hence cannot be prepared from this reaction.

Entropy change = Entropy of CaO+ entropy change of CO2- entropy change of CO2

= 213.6- (39.7+92.9) =81 J/K

Similarly enthalpy change = -393.5-(-1206.9-635.1) =1448.5 KJ

Gibbs free energy change = deltaH- TdeltaS= 1448.5-298*81/1000 =1424.362 Kj

Since the Gibbs free energy change is +ve, the reaction is not spontaneous

3. for the reaction, 2NH3+CO2--->NH2CONH2

K= [NH2CONH2]/ [PNH3]2[PCO2] and deltaG= deltaG0+RT lnK

for the 1st conditions, K = 1/(103)= 0.001, lnK= -6.91

deltaG= =13.6*1000-6.91*8.314*298 = -30720.02 Kj= -30.720 Kj. since deltaG is-ve, the reaction is product favored

for the 2nd casem deltaG= -13.6*1000+ln(1/(0.1)3* 298*8.314= 3514 Joules =3.514 Kj

The reaction is reactants favored. since deltaG is +ve.

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