Respuesta :
Answer: The value of [tex]K_{eq}[/tex] is [tex]4.84\times 10^{-5}[/tex]
Explanation:
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:
[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, concentration of ammonia = [tex]\frac{0.0280}{1.00}=0.0280M[/tex]
Concentration of oxygen gas = [tex]\frac{0.0120}{1.00}=0.0120M[/tex]
The given chemical equation follows:
[tex]4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)[/tex]
Initial: 0.0280 0.0120
At eqllm: 0.0280-4x 0.0120-3x 2x 6x
We are given:
Equilibrium concentration of nitrogen gas = [tex]3.00\times 10^{-3}M=0.003[/tex]
Evaluating the value of 'x', we get:
[tex]\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M[/tex]
Now, equilibrium concentration of ammonia = [tex]0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M[/tex]
Equilibrium concentration of oxygen gas = [tex]0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M[/tex]
Equilibrium concentration of water = [tex]6x=(6\times 0.0015)]=0.009M[/tex]
The expression of [tex]K_{eq}[/tex] for the above reaction follows:
[tex]K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}[/tex]
Hence, the value of [tex]K_{eq}[/tex] is [tex]4.84\times 10^{-5}[/tex]
