A student ran the folllowing reaction in the laboratory at 661 K: 2NH3(g) N2(g) + 3H2(g)...

A student ran the folllowing reaction in the laboratory at 661 K:

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

When she introduced NH3(g) at a pressure of 0.597 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of H2(g) to be 0.879 atm.

Calculate the equilibrium constant, Kp, she obtained for this reaction.

Expert Solution

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

I 0.597 0 0

C -2*0.293 0.293 3*0.293

E 0.011 0.293 0.879

PNH3 = 0.011atm

PN2 = 0.293atm

PH2 = 0.879atm

Kp = PN2 *P^3H2/P^2NH3

= 0.293*(0.879)^3/(0.011)^2 = 1644.55 >>>answer

queen_honey_blossom answered 2 years ago

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