At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction.

2 NH3(g) N2(g) + 3 H2(g)
At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?

(NH3)@ start = 4.0/2L = 2.0M

(NH3)@ end = 2.0mol/2L = 1M
...........2NH3 ==> N2 + 3H2
initial....2.0......0.....0
change.....-2x......x.....3x
equil.....2-2x=1.....x.....3x
For NH3 at end, 2-2x = 1 and x = 1/2; then N2 = 1/2 and H2 = 1.5
Substitute and solve for K.

1.6875

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1.69 sucka

To find the value of equilibrium constant (K) for this reaction, we need to use the balanced equation and the given information.

The balanced equation is:
2 NH3(g) ⇌ N2(g) + 3 H2(g)

At equilibrium, we are given that 2.0 mol NH3 remains. This means that 4.0 - 2.0 = 2.0 mol of NH3 has dissociated.

From the balanced equation, we can see that the mole ratio of NH3 to N2 is 2:1. This means that for every 2 mol of NH3 that dissociates, 1 mol of N2 is formed. Therefore, 1 mol of N2 is formed in this reaction.

Similarly, the mole ratio of NH3 to H2 is 2:3. This means that for every 2 mol of NH3 that dissociates, 3 mol of H2 is formed. Therefore, 3 mol of H2 is formed in this reaction.

Now we can construct the equilibrium expression using the balanced equation:
K = [N2] * [H2]^3 / [NH3]^2

Substituting the known values:
K = (1)(2^3) / (2^2)

Simplifying, we get:
K = 8 / 4
K = 2

Therefore, the value of K for this reaction is 2.