Hydrazine, H_2NNH_2, is used as a rocket fuel. Use the data below to determine i
ID: 1072799 • Letter: H
Question
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
1.
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.
2.
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. since the reaction is non spontaneous, deltaG is +Ve.
deltaG= deltaH-TdeltaS
deltaH= deltaG+ TdeltaS, deltaH is +ve since both deltaG and deltaS are +ve. Hence the reaction is endothermic.
deltaH is minimum if deltaG=0
deltaH =T*deltaS= (50+273)*104 Joules=33592 Joules= 33.592 KJ
4. deltaG= deltaGO+ RTln K
K for the reaction = [NH2CONH2] /[PNH3]2 [PCO2]= 1/(10*10*10)=0.001
where PNH3 and PCO2 are partial pressures of NH3 and CO2 respectively.
lnK= -6.91, deltaG= -13.6*1000-6.91*8.314*298 =-30720.02 Joules= -30.720 KJ,
more product is formed
for the 2nd condition, K= 1/(0.1)3= 1000,lnK=6.91
deltaG= -13.6*1000+6.91*8.314*298 =3520 J =3.520 Kj
since deltaG is +ve, the reaction is reactant favored. less product is formed.
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