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chaseafer · Feb 23, 2010, 07:44 PM

Consider the chemical reaction 2NH3(g) Picture N2(g) + 3H2(g). The equilibrium is to be established in a 1.0 L container at 1,000 K, where Kc = 4.0 × 10−2. Initially, 1,220 moles of NH3(g) are present.

Estimate the equilibrium concentration of N2(g) in mol/L. [Enter only a decimal number.]

Unknown008 · Feb 24, 2010, 07:56 AM

$2NH_3 \rightarrow 3H_2 + N_2$

The expression for Kc is:

$K_c = \frac{(P_{N_2})(P_{H_2})^3}{(P_{NH_3})^2}$

You need to find the partial pressures of NH3, N2 and H2.

That of NH3 is 1220 mol/L initially and both H and N2 are 0.

At equilibrium an unknown pressure of N2 and H2 will be settled, let's call this the partical pressure of N2 to be 'x mol/L'. Therefore, that of H2 will be 3x mol/L.
So, the NH3 left will be (1220 - 2x) mol/L.

Plug those partial pressures and the Kc value in the Kc expression:

$4 \times 10^{-2} = \frac{(x)(3x)^3}{(1220-2x)^2}$

And solve for x to get the partial pressure of N2. From that, you can convert it into concentration.