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Consider the reaction NH4Cl(aq)NH3(g) + HCl(aq) The standard free energy change for this reaction is 62.8 kJ.
The free energy change when 1.72 moles of NH4Cl(aq) react at standard condition is____ kJ.
What is the maximum amount of useful work that the reaction of 1.72 moles of NH4Cl(aq) is capable of producing in the surroundings under standard conditions?_______
If no work can be done, enter none. kJ
standard free energy change of a reaction is defined as the difference in free energy of products and reactants,when 1 mole of product is formed from the reactants while keeping both reactants and products at their standard states i.e. 1 atm pressure and 298 K temperature.It is denoted by Gr0
Thus for the reaction NH4Cl(aq) -----> NH3(g) + HCl(aq) , it is given that Gr0= 62.8 kJ, when 1 mole of each reactant react to form 1 mole of products. Since it is an extensive property , it will depend on amount of substance .
Thus for 1.72 moles taken, Gr0= 1.72*62.8 kJ
= 108.016 kJ
Part(b) Ans- None
Gibbs free energy is the maximum amount of extractable work from a system.Now since G, gibbs free energy is given by
G= H-TS (H- Enthalpy, S- Entropy)
differentiating both sides, we get
dG= dH - (T.dS +S.dT)
dG= dH - T.dS -S.dT..........................(1)
Since the given conditions are standard state, thus pressure and temperature are constant.
Now since H= U+ PV.Thus, differentiating it, we get
dH= dU+ P.dV+ V.dP.......................................(2)
Putting 2 in equation 1 and setting dP,dT=0( P,T= constant, due to standard state constraint)
dG= dU+P.dV-T.dS and from second law, dQ= T.dS. and from first law, dU=dQ-P.dV
Thus putting them in above relation , we get dG= dQ-P.dV+P.dV-dQ
dG= 0.
Thus we conclude that at constant temperature and pressure, which happens due to state of system being fixed to standard state. Thus dG, gibbs energy change which gives us maximum extractable work will be 0 or none in this case and we conclude that we can't extract any work while maintaining temperature and pressure constant.