<u>Answer:</u> The value of
is 
<u>Explanation:</u>
We are given:
Initial moles of ammonia = 0.0280 moles
Initial moles of oxygen gas = 0.0120 moles
Volume of the container = 1.00 L
Concentration of a substance is calculated by:

So, concentration of ammonia = 
Concentration of oxygen gas = 
The given chemical equation follows:

<u>Initial:</u> 0.0280 0.0120
<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x
We are given:
Equilibrium concentration of nitrogen gas = 
Evaluating the value of 'x', we get:

Now, equilibrium concentration of ammonia = ![0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M](https://tex.z-dn.net/?f=0.0280-4x%3D%5B0.0280-%284%5Ctimes%200.0015%29%5D%3D0.022M)
Equilibrium concentration of oxygen gas = ![0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M](https://tex.z-dn.net/?f=0.0120-3x%3D%5B0.0120-%283%5Ctimes%200.0015%29%5D%3D0.0075M)
Equilibrium concentration of water = ![6x=(6\times 0.0015)]=0.009M](https://tex.z-dn.net/?f=6x%3D%286%5Ctimes%200.0015%29%5D%3D0.009M)
The expression of
for the above reaction follows:
![K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BH_2O%5D%5E6%5Ctimes%20%5BN_2%5D%5E2%7D%7B%5BNH_3%5D%5E4%5Ctimes%20%5BO_2%5D%5E3%7D)
Putting values in above expression, we get:

Hence, the value of
is 