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Polkadot45 · Mar 13, 2011, 05:28 PM

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

Unknown008 · Mar 14, 2011, 04:18 AM

$2NH_3\ \rightleftharpoons\ N_2\ +\ 3H_2$

$K_c = 4.0\times10^{-2}$

Can you write down the expression for Kc?

Then, assume that x moles of NH3 dissociates to form N2 and H2. If x moles of NH3 dissociates, you'll gets x/2 moles of N2 and 3x moles of H2. From there, you get a final number of moles of NH3 as 1220-x moles.

Then, you use the expression I told you to get above and substitute in values, before solving for x, then, looking for x/2.

Can you post what you get? :)

DrBob1 · Mar 17, 2011, 02:38 PM

As hard as it is to read, the question must be dealing with 1.200 moles of NH3, not 1,200 moles. 1.2 Moles is 20.4 grams. 1,200 mols is 20.4 Kg - a little much for a 1 liter container.