Conside.. ollowing equilbrium: Now suppose a reaction vessel is filled with 7.27

Question

Conside.. ollowing equilbrium: Now suppose a reaction vessel is filled with 7.27 atm of nitrogen (N2) and 6.95 atmi of ammonia (NH,) at 854. "C. Answer the followving questions about this system: Under these conditions, will the pressure of N, tend to rise or fall? rise fall Is it possible to reverse this tendency by adding H,? In other words, if you said the pressure of N2 will tend to rise, can that be changed to a tendency to fall by adding H, Similarly, i you said the pressure of N2 will tend to fall,can that be changed to a tendency to rise by adding H2 yes the minimum pressure of H, needed to reverse it. Round your answer to 2 significant digits. 1a Check Type here to search

Explanation / Answer

1)

Answer

Yes

Explanation

N2(g) + 3H2(g) <-------> 2NH3(g)

G° = 34kJ/mol

G° = - RTlnK

lnK = -G°/RT

logK = -G°/2.303RT

= -(34000J/mol/2.303×8.314(J/mol K)×1127K)

= -34000/21579

= -1.58

K = 0.026

Q=( PNH3)2/(PN2)*(PH2)3

= (6.95atm)2/((7.27atm) * 0)

= infinity

Q > K

so, the equillibrium will shift to reactant side and the pressure of N2 will raise

2)

Answer

Yes

Explanation

By addind H2, we bring the condition Q<K, so, the equillibrium will shift product side and N2 pressure will fall

3)

Answer

6.35atm

Explanation

If we equate Q=K

0.026atm-2 =( PNH3)2/((PN2)×(PH2)3)

(PH2)3 =( PNH3)2/(PN2×0.026)

= (6.95atm)2/((7.27atm) × 0.026atm-2)

= 255.54atm3

PH2 = 6.35atm

  

  

  

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