The reaction shown below has Kp = 5.1 × 10-4. N2(g) + 3H2(g) 2NH3(g) If PN2 = 27

Question

The reaction shown below has Kp = 5.1 × 10-4.

N2(g) + 3H2(g) 2NH3(g)

If PN2 = 27.14 atm, PH2 = 33.34 atm, and PNH3 = 24.28, what would you expect to happen to the partial pressure of nitrogen as the reaction proceeds?

A. The pressure of nitrogen will stay the same as Q = K therefore the system is at equilibrium

B. The pressure of nitrogen will increase because Q < K so the reaction must proceed in reverse to reach equilibrium

C. The pressure of nitrogen will increase because Q > K so the reaction must proceed in reverse to reach equilibrium

D. The pressure of nitrogen will decrease because Q < K so the reaction must proceed forwards to reach equilibrium

E. The pressure of nitrogen will decrease because Q > K so the reaction must proceed forwards to reach equilibrium

Explanation / Answer

for the reaction Kp= [PNH3]2/ [PN2] [PH2]3 = 5.1*10-4

Q= (24.28)2/ (27.14*33.34 = 0.000586 =5.86*10-4

Q> Kp(5.1*10-4)

as the reaction proceeds, Q should reach equilibrium value and this is possible only if Q decreases and this is possilbe only if NH3 decreases. The reaction should proceed in the reverse direction.

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