Question

Consider the following equilibrium 2NH3 (g)N2(g)+3H, Now suppose a reaction vessel is filled with 2.89 atm of amoia (NH3) and 2.79 atm of nitrogen (N2) at 944. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of N2 tend to rise or fall? fall Is it possible to reverse this tendency by adding H2? In other words, if you said the pesure of N2 will tend to rise, can that be changed to a tendency to fall by adding H2? Similarly, if you said the pressure of N2 will tend to fall, can that be changed to a tendency to rise by adding H2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H2 needed to reverse it. Round your answer to 2 significant digits. yes no atm

Answer 1

given for the reaction

2NH3(g) <------> N2(g)+3H2(g), deltaG= 34 Kj= 34*1000= 34000 Kj

deltaG= -RT lnK, K= equilibrium constant, T= 944 deg.c= 944+273 K=1217K, R= gas constant= 8.314 J/mole.K

lnK= -deltaG/RT= -34*1000/(1217*8.314)= -3.36, K= 0.035

Given initial pressure of NH3= 2.89atm and that of N2= 2.79atm, PH2=0

Q= reaction coefficient = PN2*PH2³/(PNH3)²=0 <K, so the reaction proceeds forward leading to increased pressure of N2 and H2. by addition of H2, the partial pressure of H2 can be increased.

for the reaction to proceeds backward, the Q>K, so this is posssible only if

2.79*PH2³/(2.89*2.89)= 0.035 has to be used to calculate the partial pressure of H2

PH2= 0.47 for Q=K, for Q>K, so PH2>0.47 atm, hence PH2 is partial pressure of H2 need to be a minimum of 0.47 atm.