A closed system initially containing nitrogen at a pressure of 0.7900 atm and hy

Question

A closed system initially containing nitrogen at a pressure of 0.7900 atm and hydrogen at 0.4990 atm at 750 K is allowed to reach equilibrium. The value for the equilibrium constant, Kp for the reaction:

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

at 750 K is 2.79×10-5. What are the equilibrium pressures of N2, H2 and NH3 in atm?

x =    (if using the quadratic equation - only report the root that gave reasonable concentrations)

[N2]eq =    M

[H2]eq = M

[NH3]eq = M

Was the quadratic formula necessary for this problem?

Check: The x is waht % of the initial hydrogen pressure?

I have no idea where to start!

Explanation / Answer

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

Kp = [NH3]^2 / [H2]^3 [N2]

2.79X10^-5 = [NH3]^2 / [ 0.4990]^3 [0.79]

4.3X10^-4 = [NH3]^2 / (0.1242514) [0.79]

4.3X10^-4 = [NH3]^2 / (0.09815)

[NH3]^2 = 0.0000422045
[NH3] = 0.00649649905

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