Equilibrium....?

Oddveig · science forum beginner Joined: 11 Jan 2006 Posts: 4

I have the following equation:
2NH3 <-->N2+3H2

Starting with an empty container,adding 0,500mol NH3,at equilibrium we've
got left 0,100mol NH3.
What is the equilibriumconstant for this reaction?

When I'm doing the equation,I'm getting 0,400
The correct answer is however 0,423.
What am I doing wrong?

Oscar Lanzi III · science forum Guru Wannabe Joined: 30 Apr 2005 Posts: 176

You consume 0.400 mol NH3 to go down from 0.500 mol to 0.100 mol. So
how many moles of H2 do you form, and how many moles of N2 do you form?

Now put those numbers into the expression for the equilibrium constant.
Here's a hint: when I do it I get NEITHER 0.400 nor 0.423. Methinks
there is a typo somewhere.

--OL

Lasse Murtomäki · science forum beginner Joined: 17 Jun 2005 Posts: 35

Initial amount of NH3 = n0 = 0.5 mol
At equilibrium the amount of NH3 = n0 - 2x = 0.1 mol => x = 0.2 mol
At equilibrium the amount of H2 = 3x = 0.6 mol
At equilibrium the amount of N2 = x = 0.2 mol

K = [0.2*(0.6)^3]/(0.1)^2 = 4.32

If the equilibrium constant is defined as I did above.
--
Dr. Lasse Murtomäki
Helsinki University of Technology
lasse.murtomaki@tkk.fi

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